# What is Telescoping series: Definition and 18 Discussions

In mathematics, a telescoping series is a series whose general term

t

n

{\displaystyle t_{n}}
can be written as

t

n

=

a

n

a

n
+
1

{\displaystyle t_{n}=a_{n}-a_{n+1}}
, i.e. the difference of two consecutive terms of a sequence

(

a

n

)

{\displaystyle (a_{n})}
.As a consequence the partial sums only consists of two terms of

(

a

n

)

{\displaystyle (a_{n})}
after cancellation. The cancellation technique, with part of each term cancelling with part of the next term, is known as the method of differences.
For example, the series

n
=
1

1

n
(
n
+
1
)

{\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n(n+1)}}}
(the series of reciprocals of pronic numbers) simplifies as

n
=
1

1

n
(
n
+
1
)

=

n
=
1

(

1
n

1

n
+
1

)

=

lim

N

n
=
1

N

(

1
n

1

n
+
1

)

=

lim

N

[

(

1

1
2

)

+

(

1
2

1
3

)

+

+

(

1
N

1

N
+
1

)

]

=

lim

N

[

1
+

(

1
2

+

1
2

)

+

(

1
3

+

1
3

)

+

+

(

1
N

+

1
N

)

1

N
+
1

]

=

lim

N

[

1

1

N
+
1

]

=
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}}}

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1. ### How is this a telescoping series?

Hello ! Consider this series; $$\sum_{k=1}^{\infty} \frac{1}{(2k-1)(2k+1)}$$ It is said to find the limit of the series when approaches infinity.Now it is said that this is a telescopic series and that the limit is ##\frac{1}{2}## but I don't see it. I've split the an part (I don't know how...
2. ### I Divergence/Convergence for Telescoping series

Can I use the divergence test on the partial sum of the telescoping series? Lim n>infinity an if not equal zero then it diverges The example below shows a telescoping series then I found the partial sum and took the limit of it. My question is shouldn't the solution be divergent? Since the...
3. ### Telescoping Series theorem vs. Grandi's series

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4. ### Showing the sum of this telescoping series

Homework Statement Determine whether each of the following series is convergent or divergent. If the series is convergent, find its sum \sum_{i=1}^{\infty} \frac{6}{9i^{2}+6i-8} Homework Equations Partial fraction decomposition \frac{1}{3i-2} - \frac{1}{3i+4} The Attempt at a Solution...
5. ### Telescopic sum issues, cant get Sk

1. Homework Statement Find a formula for the nth partial sum of the series and use it to find the series' sum if the series converges Homework Equations 3/(1*2*3) + 3,/(2*3*4) + 3/(3*4*5) +...+ 3/n(n+1)(n+2) The Attempt at a Solution the first try, i tried using partial fraction which equals...
6. ### Why an array of telescopes is used?

To increase the resolution of an instrument, smaller wavelength and larger aperture is desirable. It is mentioned in some textbooks that the "effective" diameter of a telescope can be increased by using arrays of smaller telescopes. I just wonder why it is possible because every telescope is...
7. ### Telescoping series convergence question

Homework Statement [/B] Hello, this problem is from a well-known calc text: Σ(n=1 to ∞) 8/(n(n+2)Homework Equations [/B] What I have here is decomposingg the problem into Σ(n=1 to ∞)(8/n -(8/n+2)The Attempt at a Solution I have the series sum as equaling (8/1-8/3) + (8/2-8/4) + (8/3-8/5) +...
8. ### Help with Telescoping Series Problem

Homework Statement Problem is attached in this post.Homework Equations Problem is attached in this post. The Attempt at a Solution Attempt at solution is attached. Apparently the answer is -pi/6, however I've solved for it several times and keep getting my answer as pi/3 via Telescoping Series
9. ### Partial Sum Formula of Telescoping Series: 1/2*(1+1/2-1/(n+1)-1/(n+2))

Homework Statement Find the formula of the partial sum of the series Ʃ1/[k(k+2)] k from 1 to infinity Homework Equations The Attempt at a Solution Using partial fractions i rewrite the series as 1/2*Ʃ[1/k] - [1/(k+2)] Then I start writing out the series from k=1 to 5...
10. ### Telescoping Series involving 3 parts.

Homework Statement Determine whether the series is convergent or divergent by expressing s sub n as a telescoping series. If convergent find the sum. \sum_{n=2}^\infty \frac{1}{n^3-1} The Attempt at a Solution First thing I did was partial fraction decomposition. Resulting in: Ʃ (n=2 to ∞)...
11. ### Interesting Telescoping Series - Calc 2

Interesting Telescoping Series -- Calc 2 My problem with this series arose when I attempted a partial fractions decompisition of the following, (k-1)/(2^(2k+1). I attempted to factor the denominator with 2(2^k) which is right, but where do I go from there? It does not help much to multiply...
12. ### Understanding Telescoping Series: Finding the Sum

My apologies beforehand for not using the right format for this post. Homework Statement Find the sum of (from 1 to inf) of \sum8/(n(n+1)(n+2)) Homework Equations The Attempt at a Solution I approached the problem like I would a telescoping series by using partial faction...
13. ### Help with this Telescoping Series

Homework Statement Show that \sum_{n=1}^{\infty} \frac{1}{(n+1)(n+2)} = 1 Homework Equations The Attempt at a Solution So as this is a telescoping series, I rewrote the general formula through partial fractions as \sum_{n=1}^{\infty} \frac{1}{(n+1)} - \frac{1}{(n+2)} = 1 The first few...
14. ### Understanding telescoping series

I understand that if given the following series: \sum _{n=1}^{\infty } \frac{1}{n(n+3)} I can break it up using partial fraction decomposition into: \sum _{n=1}^{\infty } \frac{1}{3n}-\frac{1}{3n+9} If I start listing the values of n_1,n_2, etc, I can see the cancellation pattern and find...
15. ### Convergent Telescoping Series: \sum_{n=1}^{\infty} \arctan(n+1)-\arctan(n)

\sum_{n=1}^{\infty} \arctan(n+1)-\arctan(n) . So we want to take \lim s_{n} = \lim_{n\rightarrow \infty} (\arctan 2-\arctan 1)+(\arctan 3-\arctan 2) + ... + (\arctan(n+1)-\arctan n) . From this I can see all of the terms cancel except \arctan 1 . But then how do we get: \lim_{n\rightarrow...
16. ### Telescoping Series Simplification

I want to determine whether \sum^{\infty}_{n=2} \frac{2}{n^{2}-1} is convergent or divergent. I did the following: \sum^{\infty}_{n=2} \frac{2}{n^{2}-1} = \sum^{\infty}_{n=2} (\frac{1}{j-1} + \frac{1}{j+1}) . Writing down some terms, I got...
17. ### Solving Telescoping Series with n Terms

I have no idea how to do this :'( really the only part I don't understand is the ending part...like for the infinite sum of (1/n(n+1))....I know you start of by partial fractions...then you just plug in a few numbers for n. so I end up with: (1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) + ...
18. ### Quick question about a telescoping series

\sum_{j=k}^{\infty}\left\{{\frac{1}{b_{j}}-\frac{1}{b_{j+1}}}\right\}=\frac{1}{b_{k}} This series holds for all real monotone sequences of b_j . So if I were to carry this series out to say n I end up with a partial sum that looks like...