# nth partial sums of series + telescoping series

• Apr 12th 2011, 01:32 PM
Wolvenmoon
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!
• Apr 12th 2011, 08:50 PM
NOX Andrew
The series $\displaystyle \displaystyle 2 + \dfrac{2}{3} + \dfrac{2}{9} + ...$ is geometric.

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

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

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

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

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

Subtract equation 2 from equation 1.

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

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

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

Now, it's a matter of recognizing geometric series and identifying $\displaystyle \displaystyle a_0$ and $\displaystyle \displaystyle r$. What is $\displaystyle \displaystyle a_0$ and $\displaystyle \displaystyle r$ for the geometric series $\displaystyle \displaystyle 2 + \dfrac{2}{3} + \dfrac{2}{9} + ...$?
• Apr 12th 2011, 08:50 PM
matheagle
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.

$\displaystyle S_N=a_1+\cdots +a_N$

AND its LIMIT is the infinite sum

$\displaystyle \sum_{i=1}^{\infty}a_i=\lim_{N\to\infty}\sum_{i=1} ^{N}a_i =\lim_{N\to\infty}S_N$
• Apr 12th 2011, 11:38 PM
NOX Andrew
The series $\displaystyle \displaystyle \sum\limits_{n = 1}^{\infty}\left(\dfrac{1}{n} - \dfrac{1}{n + 1}\right)$ is a telescoping series.

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

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

Simplify the right-hand side of the equation.

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

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

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

All but the first and last terms cancel.

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

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

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

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

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

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.
• Apr 13th 2011, 11:39 AM
Wolvenmoon
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.