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
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...
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...
Homework Statement
No actual problem, thinking about the telescoping series theorem and Grandi's series
For reference Grandi's series S = 1 - 1 + 1 - 1...
Homework Equations
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The telescoping series theorem in my book states that a telescoping series of the form (b1 - b2) + ... + (bn -...
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...
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...
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...
Homework Statement
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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) +...
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
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...
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 ∞)...
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...
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...
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...
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...
\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...
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...
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) + ...
\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...