# Thread: nth partial sums of series + telescoping series

1. ## nth partial sums of series + telescoping series

Alright, so I'm having some conceptual trouble. I'll go ahead and post the homework problems but these are the concept in practice. I don't need all of the homework worked. Just pick and choose and help me see what's going on here. I only need to figure out what the nth partial sum is and how to reliably find it.

Homework problems, prefer help on one or two of the latter 3 but the first 2 are easier if you don't have the time.

"Find a formula for the nth partial sum of each series and use it to find the series' sum if the series converges."

2+ 2/3 + 2/9 + 2/27 + .... 2/3^(n-1) + ...

(Answer: s_n=((2(1-(1/3)^n)) / (1-(1/3) and it converges at 3.)
and

1/(2*3) + 1/(3*4) + 1/(4*5) + ... 1/((n+1)(n+2))

(Answer is: s_n=1/2 - 1/(n+2) converges at 1/2)

Then part 2 where I got my butt kicked for not understanding partial sums, in the homework,

"Find a formula for the nth partial sum of the series and use it to determine if the series converges or diverges. If it converges, find the sum."

Note: I only need help finding s_n right now!

Summation from n = 1 to infinity of (1/n - 1/(n+1)
( Answer: s_n= 1 - 1/n+1, converges at 1 )

and

Summation from n=1 to infinity of ln((n+1)^(1/2)) - ln(n^(1/2))

( Answer: s_n= ln((n+1)^(1/2)), diverges )

and

Summation from n=1 to infinity of 4/((4n-3)(4n+1))

Thanks guys! Pick and choose! Once I get it nailed down on one, the others will come easily!

2. The series $\displaystyle 2 + \dfrac{2}{3} + \dfrac{2}{9} + ...$ is geometric.

I will demonstrate how to find the sum of the first $\displaystyle n$ terms of any geometric sequence where $\displaystyle a_0$ is the first term of the sequence and $\displaystyle r$ is the ratio of the sequence.

The sum of the first $\displaystyle n$ terms, $\displaystyle S_n$, is given by

$\displaystyle S_n = a_0 + a_0 r + a_0 r^2 + a_0 r^3 + ... a_0 r^n$ (1)

Multiply both sides of the equation by $\displaystyle r$.

$\displaystyle r S_n = a_0 r + a_0 r^2 + a_0 r^3 + a_0 r^4 + ... a_0 r^{n+1}$ (2)

Subtract equation 2 from equation 1.

$\displaystyle S_n - r S_n = a_0 - a_0 r^{n + 1} \iff S_n\left(1 - r\right) = a_0\left(1 - r^{n + 1}\right)$

Divide both sides of the equation by $\displaystyle 1 - r$.

$S_n = \dfrac{a_0 \left(1 - r^{n + 1}\right)}{1 - r}$

Now, it's a matter of recognizing geometric series and identifying $\displaystyle a_0$ and $\displaystyle r$. What is $\displaystyle a_0$ and $\displaystyle r$ for the geometric series $\displaystyle 2 + \dfrac{2}{3} + \dfrac{2}{9} + ...$?

3. you mean "converges to 3"

I think what you need to understand is what a partial sum is.
It is the sum of the first N terms.

$S_N=a_1+\cdots +a_N$

AND its LIMIT is the infinite sum

$\sum_{i=1}^{\infty}a_i=\lim_{N\to\infty}\sum_{i=1} ^{N}a_i =\lim_{N\to\infty}S_N$

4. The series $\displaystyle \sum\limits_{n = 1}^{\infty}\left(\dfrac{1}{n} - \dfrac{1}{n + 1}\right)$ is a telescoping series.

The sum of the first $\displaystyle N$ terms of a telescopic sequence, $\displaystyle S_N$, is given by

$\displaystyle S_N = \left(\dfrac{1}{1} - \dfrac{1}{1 + 1}\right) + \left(\dfrac{1}{2} - \dfrac{1}{2 + 1}\right) + \left(\dfrac{1}{3} - \dfrac{1}{3 + 1}\right) + \left(\dfrac{1}{4} - \dfrac{1}{4 + 1}\right) + ... + \left(\dfrac{1}{N} - \dfrac{1}{N + 1}\right)$

Simplify the right-hand side of the equation.

$\displaystyle S_N = \left(1 - \dfrac{1}{2}\right) + \left(\dfrac{1}{2} - \dfrac{1}{3}\right) + \left(\dfrac{1}{3} - \dfrac{1}{4}\right) + \left(\dfrac{1}{4} - \dfrac{1}{5}\right) + ... + \left(\dfrac{1}{N} - \dfrac{1}{N + 1}\right)$

Apply the associative property of addition to conveniently rearrange the terms.

$\displaystyle S_N = 1 + \left(-\dfrac{1}{2} + \dfrac{1}{2}\right) + \left(-\dfrac{1}{3} + \dfrac{1}{3}\right) + \left(-\dfrac{1}{4} + \dfrac{1}{4}\right) + ... + \left(-\dfrac{1}{N} + \dfrac{1}{N}\right) - \dfrac{1}{N + 1}$

All but the first and last terms cancel.

$\displaystyle S_N = 1 - \dfrac{1}{N + 1}$

A general telescopic series looks like $\displaystyle \sum\limits_{n = 1}^{\infty} \left(a_n - a_{n + 1}\right)$. The sum of the first $\displaystyle N$ terms of a telescopic series, $\displaystyle S_N$, is given by

$\displaystyle S_N = a_1 - a_{N + 1}$

In words, the sum of the first $\displaystyle N$ terms of a telescopic sequence is the sum of the first term and the $\displaystyle (N + 1)$th term (because all the terms in between cancel).

Like before, it's now a matter of recognizing telescopic series and identifying $\displaystyle a_1$ and $\displaystyle a_{N + 1}$.

The remaining series are telescopic - algebraic manipulation helps to recognize them as much. Don't hesitate to ask any more questions. However, I suggest reviewing the section on series in your textbook as all of your problems could have been solved with knowledge of common series formulas.

5. Didn't see your post, Andrew, I'd left it up overnight! Will double reply soon.

Original post:
Oops @ the geometric series. We were not shown that s_n nor did he work a problem like this. He pretty much dropped summation n to infinity of a * r^n-1 = a/1-r if |r|<1 and moved on.

So what is the quick and dirty way to always find the nth partial sum so that I can take the limit to see where a series converges? Both when given a set of numbers and when given an actual series?

It's likely he's going to quiz tonight regardless of how many people in the class are baffled, so I need to get this patched up and figured out.

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### 4/(4n-3) (4n 1) using telescoping series

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