Series - nth term - Is this right?

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October 25th 2008, 08:10 PM
tiredNConfused

Series - nth term - Is this right?

I am not 100% confident about this so would someone mind checking my work?

The nth partial sum of a Series is given by:
$Sn=\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+...+\ frac{1}{2n}=\sum_{k=1}^{n}\frac{1}{n+k}$

1. What is the associated Series:

$\sum_{n=1}^{\infty}\frac{1}{2n}$

2. Convergence?

$\lim_{n\to\infty}\frac{1}{2n}=0$

General term goes to 0

Test for convergence by direct comparison.
$\frac{1}{2n}=\frac{1}{n-\frac{1}{2}}>\frac{1}{n}$

$\frac{1}{n}$ divergent harmonic series

$\sum_{n=1}^{\infty}\frac{1}{2n}$ diverges by direct comparison with $\frac{1}{n}$

3. Determine the Limit if possible

$\lim_{n\to\infty}\;Sn=\lim_{n\to\infty}\frac{1}{2n }=0$
October 25th 2008, 08:19 PM
ThePerfectHacker

Quote:

Originally Posted by tiredNConfused

I am not 100% confident about this so would someone mind checking my work?

The nth partial sum of a Series is given by:
$Sn=\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+...+\ frac{1}{2n}=\sum_{k=1}^{n}\frac{1}{n+k}$

1. What is the associated Series:

Let $a_n = \left\{ \begin{array}{c} \frac{1}{2} \text{ for }n=1 \\ \frac{1}{2n-1}+\frac{1}{2n} \text{ for }n\geq 2 \end{array} \right.$

Then the series is $\sum_{n=1}^{\infty} a_n$

This series converges if and only if $\sum_{n=2}^{\infty} a_n = \sum_{n=2}^{\infty} \left( \frac{1}{2n-1} + \frac{1}{2n} \right)$ converges.

Can you finish this now?
October 25th 2008, 08:25 PM
tiredNConfused

Yes I can, I knew it wasn't as easy as I made it.

But one question, how did you come up with: $\frac{1}{2n-1}$
October 25th 2008, 08:35 PM
ThePerfectHacker

Quote:

Originally Posted by tiredNConfused

Yes I can, I knew it wasn't as easy as I made it.

But one question, how did you come up with: $\frac{1}{2n-1}$

Because,
$s_1 = \boxed{a_1 = \tfrac{1}{2}}$
$s_2 = \boxed{a_1+a_2 = \tfrac{1}{2}+\tfrac{1}{3}+\tfrac{1}{4}}$
$s_3 = \boxed{a_1+a_2+a_3 = \tfrac{1}{2}+\tfrac{1}{3}+\tfrac{1}{4}+\tfrac{1}{5 }+\tfrac{1}{6}}$

Now solve the first boxed equation, well it is already solved, $a_1 = \tfrac{1}{2}$ .

For the second boxed equation we have $a_1 + a_2 = \tfrac{1}{2}+\tfrac{1}{3}+\tfrac{1}{4}$ and we know what $a_1$ is.
Solving it gives, $a_2 = \tfrac{1}{3}+\tfrac{1}{4}$ .

Similar with the third boxed equation we get $a_3 = \tfrac{1}{5} + \tfrac{1}{6}$

And this generalizes. That is how we get the formula for $a_1$ .
October 25th 2008, 08:36 PM
tiredNConfused

Thanks soooo much!!!