Telescoping series

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In mathematics, a telescoping series is a series whose general term ${\displaystyle t_{n}}$ is of the form ${\displaystyle t_{n}=a_{n+1}-a_{n}}$, i.e. the difference of two consecutive terms of a sequence ${\displaystyle (a_{n})}$. As a consequence the partial sums of the series only consists of two terms of ${\displaystyle (a_{n})}$ after cancellation.^[1][2]

The cancellation technique, with part of each term cancelling with part of the next term, is known as the method of differences.

An early statement of the formula for the sum or partial sums of a telescoping series can be found in a 1644 work by Evangelista Torricelli, De dimensione parabolae.^[3]

Telescoping sums are finite sums in which pairs of consecutive terms partly cancel each other, leaving only parts of the initial and final terms.^[1][4] Let ${\displaystyle a_{n}}$ be the elements of a sequence of numbers. Then $${\displaystyle \sum _{n=1}^{N}\left(a_{n}-a_{n-1}\right)=a_{N}-a_{0}.}$$ If ${\displaystyle a_{n}}$ converges to a limit ${\displaystyle L}$, the telescoping series gives: $${\displaystyle \sum _{n=1}^{\infty }\left(a_{n}-a_{n-1}\right)=L-a_{0}.}$$

Every series is a telescoping series of its own partial sums.^[5]

• Using the property that the square of integer A is the sum of the first A odd integers, or:^[6]

${\displaystyle A^{2}=\sum _{i=1}^{A}2i-1}$

And then expanding the telescoping sum we have:

${\displaystyle \sum _{i=1}^{a}{(2i-1)}=a^{2}-{\cancel {(a-1)^{2}}}+{\cancel {(a-1)^{2}}}-{\cancel {(a-2)^{2}}}+{\cancel {(a-2)^{2}}}-...+{\cancel {1}}-{\cancel {1}}=a^{2}}$

So for example ${\displaystyle a=5}$  ; ${\displaystyle a^{2}=1+3+5+7+9=25}$

From where the general formula for any n-th power of an integer A:

${\displaystyle A^{n}=\sum _{X=1}^{A}(X^{n}-(X-1)^{n})}$

This open a breach in the forgotten math fact that any parabola (or polynomial) subtend an area from 0 to an integer abscissa A that can be squared via integral or via a finite sum since using X instead of i, and representing what we are doing on the Cartesian plane will be immediately clear that it is possible to apply an exchange of variable ${\displaystyle x=X/K}$ let one write a Sum capable to moves ${\displaystyle Step=integer}$ ${\displaystyle index}$ * ${\displaystyle scalefactor=1/K}$ different from an integer just. So opening the way to calculus:

The new Scaling Rule (holding the same physical Area): From Sum of Integers to Sum of Rationals, then to the Limit

More in general, remembering the new definition for the Sum operator as given into the Abstract, we can first use and push the telescoping Sum properties to the limit (then talk in a modular like concept) to show How to refine a Sum, so working to have at the end not just the same numerical value, but the same physical Area (in square meter f.ex.) squareing with a finite number or rectangles called Gnomons, the Area Below the 1st Derivative of a Parabola ${\displaystyle Y=X^{n}}$ and then show the 4 following identities that are true just for an Upper Limit is ${\displaystyle A\in \mathbb {N^{+}} }$ . This property will be called ${\displaystyle K}$ invariance for plynomials.

Remembering that:

${\displaystyle A^{2}*K^{2}=\sum _{X=1}^{A\cdot K}(2X-1)}$

More in general:

${\displaystyle A^{n}*K^{n}=\sum _{X=1}^{A\cdot K}(X^{n}-(X-1)^{n})}$

Thanks to the known distributive property for the Sum we can left unchanged the value of the Sum if we multiply all the sum by an unitary (in this case quadratic) factor ${\displaystyle 1={\frac {K^{2}}{K^{2}}}}$ , reveal us a nice surprise once split into the Sum in this way:

${\displaystyle A^{2}=\left(\sum _{X=1}^{A}2X-1\right){\frac {K^{2}}{K^{2}}}=\sum _{X=1}^{A*K}{\frac {2X-1}{K^{2}}}}$

then I'll show how to apply the exchange of variable ${\displaystyle X=x*K}$ having:

${\displaystyle A^{2}=\sum _{X=1}^{A*K}{\frac {2X-1}{K^{2}}}=\sum _{x=1/K}^{\frac {A*{\cancel {K}}}{\cancel {K}}}{\frac {2x\cdot {\cancel {K}}}{K^{\cancel {2}}}}-{\frac {1}{K^{2}}}=\sum _{x=1/K}^{A}\left({\frac {2x}{K}}-{\frac {1}{K^{2}}}\right)}$

And I hope is now clear why it is used ${\displaystyle X}$ instead of ${\displaystyle i}$ as the New Step (or talking-scaled index) of such sums (as will be proved hereafter).

IF and only IF (IFF) the Upper Limit ${\displaystyle A\in \mathbb {N^{+}} }$ we can now write the following quadruple equality:

${\displaystyle A^{2}=\sum _{X=1}^{A}2X-1=\sum _{x=1/K}^{A}\left({\frac {2x}{K}}-{\frac {1}{K^{2}}}\right)=\lim _{K\to \infty }\sum _{x=1/K}^{A}\left({\frac {2x}{K}}-{\frac {1}{K^{2}}}\right)=\int _{0}^{A}2xdx=A^{2}}$

That shows what I will call: the Distribution for Power Terms Law, that works for the n-th power in this way:

${\displaystyle A^{n}=\sum _{X=1}^{A}M_{n}=\sum _{x=1/K}^{A}M_{n,K}}$

where ${\displaystyle M_{n}}$ was already presented, and ${\displaystyle M_{n,K}}$ is as follow (Pls see reference for the proof) and show the "External Factor Distributive Law for Powers". so how the scaling, so the exchange of variable, will affect the Terms of the Sum (remembering the result of the sum rest the same):

${\displaystyle M_{n,K}={n \choose 1}{\frac {x^{n-1}}{K}}-{n \choose 2}{\frac {x^{n-2}}{K^{2}}}+{n \choose 3}{\frac {x^{n-3}}{K^{3}}}-...+/-{\frac {1}{K^{n}}}}$

The first ${\displaystyle M_{n,K}}$ can be easily written remembering the Tartaglia's triangle (so the binomial develop) for ${\displaystyle (X-1)^{n}}$ for what you have after to eliminate the first term of the develop, alternatively changing the sign from - to +, to have ${\displaystyle (X^{n}-(X-1)^{n})}$:

${\displaystyle M_{2,K}={\frac {2x}{K}}-{\frac {1}{K^{2}}}}$

${\displaystyle M_{3,K}={\frac {3x^{2}}{K}}-{\frac {3x}{K^{2}}}+{\frac {1}{K^{3}}}}$

${\displaystyle M_{4,K}={\frac {4x^{3}}{K}}-{\frac {6x^{2}}{K^{2}}}+{\frac {4x}{K^{3}}}-{\frac {1}{K^{4}}}}$

${\displaystyle M_{5,K}={\frac {5x^{4}}{K}}-{\frac {10x^{3}}{K^{2}}}-{\frac {10x^{2}}{K^{3}}}+{\frac {5x}{K^{4}}}-{\frac {1}{K^{5}}}}$

etc...

Then a list of new manipulation will reveal a new way to solve Power Problems, and conditions let some equality be possible or not, so True or False (as Fermat The Last for ${\displaystyle n=2}$ and for all ${\displaystyle n<2}$).

${\displaystyle A^{n}=\sum _{X=1}^{A}(X^{n}-(X-1)^{n})=\sum _{x=1/K}^{A}({n \choose 1}{\frac {x^{n-1}}{K}}-{n \choose 2}{\frac {x^{n-2}}{K^{2}}}+{n \choose 3}{\frac {x^{n-3}}{K^{3}}}-...+/-{\frac {1}{K^{n}}})}$

This also leads to 2 new sum properties for sum of polynomials can be shownf.ex. into this equalities... it seems no mathematician want to admit...

${\displaystyle A^{3}=\sum _{1}^{A}3X^{2}-3X+1=\sum _{x=1/K}^{A}{\frac {3x^{2}}{K}}-{\frac {3x}{K^{2}}}+{\frac {1}{K^{3}}}=\sum _{x=A/B}^{B}{\frac {3Ax^{2}}{B}}-{\frac {3xA^{2}}{B^{2}}}+{\frac {A^{3}}{B^{3}}}}$

or

${\displaystyle A^{3}=\sum _{x=A/B}^{B}{\frac {3Ax^{2}}{B}}-{\frac {3xA^{2}}{B^{2}}}+{\frac {A^{3}}{B^{3}}}=\sum _{X=1}^{B}{\frac {3A}{B}}\left({\frac {A}{B}}X\right)^{2}-{\frac {3A^{2}}{B^{2}}}\left({\frac {A}{B}}X\right)+{\frac {A^{3}}{B^{3}}}}$

That onto the Right Hand of Fermat the Last let one write:

${\displaystyle C^{3}-B^{3}=\sum _{x=1/K}^{C-B}{\frac {3(x+B)^{2}}{K}}-{\frac {3(x+B)}{K^{2}}}+{\frac {1}{K^{3}}}}$

${\displaystyle =\sum _{x={\frac {C-B}{A}}}^{C-B}\left({\frac {3(C-B)(x+B)^{2}}{A}}-{\frac {3(C-B)^{2}(x+B)}{A^{2}}}+{\frac {(C-B)^{3}}{A^{3}}}\right)}$

${\displaystyle =\sum _{X={\frac {(C-B)A}{C-B}}=1}^{{\frac {(C-B)A}{C-B}}=A}{\frac {3(C-B)(X{\frac {C-B}{A}}+B)^{2}}{A}}-{\frac {3(C-B)^{2}(X{\frac {C-B}{A}}+B)}{A^{2}}}+{\frac {(C-B)^{3}}{A^{3}}}}$

so:

${\displaystyle C^{3}-B^{3}=\sum _{X=1}^{A}{\frac {3(C-B)(X{\frac {C-B}{A}}+B)^{2}}{A}}-{\frac {3(C-B)^{2}(X{\frac {C-B}{A}}+B)}{A^{2}}}+{\frac {(C-B)^{3}}{A^{3}}}}$ And also: ${\displaystyle C^{3}-B^{3}=\sum _{x={\frac {1}{K}}}^{A}{\frac {3(C-B)({\frac {C-B}{A}}x+B)^{2}}{AK}}-{\frac {3(C-B)^{2}\cdot ({\frac {C-B}{A}}x+B)}{(AK)^{2}}}+{\frac {(C-B)^{3}}{(AK)^{3}}}=}$

${\displaystyle C^{3}-B^{3}=\sum _{x={\frac {1}{A}}}^{A}{\frac {3(C-B)({\frac {C-B}{A}}x+B)^{2}}{A^{2}}}-{\frac {3(C-B)^{2}\cdot ({\frac {C-B}{A}}x+B)}{(A^{2})^{2}}}+{\frac {(C-B)^{3}}{(A^{2})^{3}}}=}$

${\displaystyle C^{3}-B^{3}=\sum _{X=1}^{A^{2}}{\frac {3(C-B)({\frac {C-B}{A^{2}}}X+B)^{2}}{A^{2}}}-{\frac {3(C-B)^{2}\cdot ({\frac {C-B}{A^{2}}}X+B)}{(A^{2})^{2}}}+{\frac {(C-B)^{3}}{(A^{2})^{3}}}}$

• The product of a geometric series with initial term ${\displaystyle a}$ and common ratio ${\displaystyle r}$ by the factor ${\displaystyle (1-r)}$ yields a telescoping sum, which allows for a direct calculation of its limit:^[7]

$${\displaystyle (1-r)\sum _{n=0}^{\infty }ar^{n}=\sum _{n=0}^{\infty }ar^{n}-ar^{n+1}=a}$$ when ${\displaystyle |r|<1,}$ so when ${\displaystyle |r|<1,}$ $${\displaystyle \sum _{n=0}^{\infty }ar^{n}={\frac {a}{1-r}}.}$$

• The series

$${\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n(n+1)}}}$$ is the series of reciprocals of pronic numbers, and it is recognizable as a telescoping series once rewritten in partial fraction form^[1] ${\displaystyle {\begin{aligned}\sum _{n=1}^{\infty }{\frac {1}{n(n+1)}}&{}=\sum _{n=1}^{\infty }\left({\frac {1}{n}}-{\frac {1}{n+1}}\right)\\{}&{}=\lim _{N\to \infty }\sum _{n=1}^{N}\left({\frac {1}{n}}-{\frac {1}{n+1}}\right)\\{}&{}=\lim _{N\to \infty }\left\lbrack {\left(1-{\frac {1}{2}}\right)+\left({\frac {1}{2}}-{\frac {1}{3}}\right)+\cdots +\left({\frac {1}{N}}-{\frac {1}{N+1}}\right)}\right\rbrack \\{}&{}=\lim _{N\to \infty }\left\lbrack {1+\left(-{\frac {1}{2}}+{\frac {1}{2}}\right)+\left(-{\frac {1}{3}}+{\frac {1}{3}}\right)+\cdots +\left(-{\frac {1}{N}}+{\frac {1}{N}}\right)-{\frac {1}{N+1}}}\right\rbrack \\{}&{}=\lim _{N\to \infty }\left\lbrack {1-{\frac {1}{N+1}}}\right\rbrack =1.\end{aligned}}}$

• Let k be a positive integer. Then

$${\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n(n+k)}}={\frac {H_{k}}{k}}}$$ where H_k is the kth harmonic number.

• Let k and m with k ${\displaystyle \neq }$ m be positive integers. Then

$${\displaystyle \sum _{n=1}^{\infty }{\frac {1}{(n+k)(n+k+1)\dots (n+m-1)(n+m)}}={\frac {1}{m-k}}\cdot {\frac {k!}{m!}}}$$ where ${\displaystyle !}$ denotes the factorial operation.

• Many trigonometric functions also admit representation as differences, which may reveal telescopic canceling between the consecutive terms. Using the angle addition identity for a product of sines,

$${\displaystyle {\begin{aligned}\sum _{n=1}^{N}\sin \left(n\right)&{}=\sum _{n=1}^{N}{\frac {1}{2}}\csc \left({\frac {1}{2}}\right)\left(2\sin \left({\frac {1}{2}}\right)\sin \left(n\right)\right)\\&{}={\frac {1}{2}}\csc \left({\frac {1}{2}}\right)\sum _{n=1}^{N}\left(\cos \left({\frac {2n-1}{2}}\right)-\cos \left({\frac {2n+1}{2}}\right)\right)\\&{}={\frac {1}{2}}\csc \left({\frac {1}{2}}\right)\left(\cos \left({\frac {1}{2}}\right)-\cos \left({\frac {2N+1}{2}}\right)\right),\end{aligned}}}$$ which does not converge as ${\textstyle N\rightarrow \infty .}$

In probability theory, a Poisson process is a stochastic process of which the simplest case involves "occurrences" at random times, the waiting time until the next occurrence having a memoryless exponential distribution, and the number of "occurrences" in any time interval having a Poisson distribution whose expected value is proportional to the length of the time interval. Let X_t be the number of "occurrences" before time t, and let T_x be the waiting time until the xth "occurrence". We seek the probability density function of the random variable T_x. We use the probability mass function for the Poisson distribution, which tells us that

${\displaystyle \Pr(X_{t}=x)={\frac {(\lambda t)^{x}e^{-\lambda t}}{x!}},}$

where λ is the average number of occurrences in any time interval of length 1. Observe that the event {X_t ≥ x} is the same as the event {T_x ≤ t}, and thus they have the same probability. Intuitively, if something occurs at least ${\displaystyle x}$ times before time ${\displaystyle t}$, we have to wait at most ${\displaystyle t}$ for the ${\displaystyle xth}$ occurrence. The density function we seek is therefore

${\displaystyle {\begin{aligned}f(t)&{}={\frac {d}{dt}}\Pr(T_{x}\leq t)={\frac {d}{dt}}\Pr(X_{t}\geq x)={\frac {d}{dt}}(1-\Pr(X_{t}\leq x-1))\\\\&{}={\frac {d}{dt}}\left(1-\sum _{u=0}^{x-1}\Pr(X_{t}=u)\right)={\frac {d}{dt}}\left(1-\sum _{u=0}^{x-1}{\frac {(\lambda t)^{u}e^{-\lambda t}}{u!}}\right)\\\\&{}=\lambda e^{-\lambda t}-e^{-\lambda t}\sum _{u=1}^{x-1}\left({\frac {\lambda ^{u}t^{u-1}}{(u-1)!}}-{\frac {\lambda ^{u+1}t^{u}}{u!}}\right)\end{aligned}}}$

The sum telescopes, leaving

${\displaystyle f(t)={\frac {\lambda ^{x}t^{x-1}e^{-\lambda t}}{(x-1)!}}.}$

For other applications, see:

• Proof that the sum of the reciprocals of the primes diverges, where one of the proofs uses a telescoping sum;
• Fundamental theorem of calculus, a continuous analog of telescoping series;
• Order statistic, where a telescoping sum occurs in the derivation of a probability density function;
• Lefschetz fixed-point theorem, where a telescoping sum arises in algebraic topology;
• Homology theory, again in algebraic topology;
• Eilenberg–Mazur swindle, where a telescoping sum of knots occurs;
• Faddeev–LeVerrier algorithm.

A telescoping product is a finite product (or the partial product of an infinite product) that can be canceled by the method of quotients to be eventually only a finite number of factors.^[8][9] It is the finite products in which consecutive terms cancel denominator with numerator, leaving only the initial and final terms. Let ${\displaystyle a_{n}}$ be a sequence of numbers. Then, $${\displaystyle \prod _{n=1}^{N}{\frac {a_{n-1}}{a_{n}}}={\frac {a_{0}}{a_{N}}}.}$$ If ${\displaystyle a_{n}}$ converges to 1, the resulting product gives: $${\displaystyle \prod _{n=1}^{\infty }{\frac {a_{n-1}}{a_{n}}}=a_{0}}$$

For example, the infinite product^[8] $${\displaystyle \prod _{n=2}^{\infty }\left(1-{\frac {1}{n^{2}}}\right)}$$ simplifies as $${\displaystyle {\begin{aligned}\prod _{n=2}^{\infty }\left(1-{\frac {1}{n^{2}}}\right)&=\prod _{n=2}^{\infty }{\frac {(n-1)(n+1)}{n^{2}}}\\&=\lim _{N\to \infty }\prod _{n=2}^{N}{\frac {n-1}{n}}\times \prod _{n=2}^{N}{\frac {n+1}{n}}\\&=\lim _{N\to \infty }\left\lbrack {{\frac {1}{2}}\times {\frac {2}{3}}\times {\frac {3}{4}}\times \cdots \times {\frac {N-1}{N}}}\right\rbrack \times \left\lbrack {{\frac {3}{2}}\times {\frac {4}{3}}\times {\frac {5}{4}}\times \cdots \times {\frac {N}{N-1}}\times {\frac {N+1}{N}}}\right\rbrack \\&=\lim _{N\to \infty }\left\lbrack {\frac {1}{2}}\right\rbrack \times \left\lbrack {\frac {N+1}{N}}\right\rbrack \\&={\frac {1}{2}}\times \lim _{N\to \infty }\left\lbrack {\frac {N+1}{N}}\right\rbrack \\&={\frac {1}{2}}.\end{aligned}}}$$