# Proving that a series diverges - help!

• Apr 28th 2013, 10:38 AM
kaya2345
Proving that a series diverges - help!
So i've been doing some work on sequences lately and got stuck on this question (it has been attached as an image) Attachment 28185

I know that I have to use the nth term test for divergence but I am not sure of how to go about doing it

any help is much appreciated! thank you
• Apr 28th 2013, 11:11 AM
Plato
Re: Proving that a series diverges - help!
Quote:

Originally Posted by kaya2345
So i've been doing some work on sequences lately and got stuck on this question (it has been attached as an image) Attachment 28185
I know that I have to use the nth term test for divergence.

You should know that ${\lim _{n \to \infty }}{\left( {1 + \frac{a}{{n + b}}} \right)^{cn}} = {e^{ac}}$

Well ${\lim _{k \to \infty }}{\left( {\frac{k}{{k + 1}}} \right)^k} = {\lim _{k \to \infty }}{\left( {1 + \frac{{ - 1}}{{k + 1}}} \right)^k}$
• Apr 28th 2013, 11:25 AM
emakarov
Re: Proving that a series diverges - help!
Alternatively, represent $\left(\frac{x}{x+1}\right)^x$ as $e^{x\ln(x/(x+1))}$ and then prove that $x\ln(x/(x+1))\to-1$ as $x\to\infty$ using either Taylor series or L'Hopital's rule.
• Apr 28th 2013, 03:32 PM
Soroban
Re: Proving that a series diverges - help!
Hello, kaya2345!

We are expected to know this definition:

. . $\lim_{n\to\infty}\left(1+\tfrac{1}{n}\right)^n \;=\;e$

Quote:

Prove that that the following series diverges: . $\sum^{\infty}_{k=1}\left(\frac{k}{k+1}\right)^k$

Theorem: If the nth term does not approach zero, the series diverges.

We have: . $a_n \;=\;\left(\frac{k}{k+1}\right)^k \;=\;\left(\frac{1}{\frac{k+1}{k}}\right)^k \;=\;\left(\frac{1}{1+\frac{1}{k}}\right)^k \;=\;\frac{1}{(1+\frac{1}{k})^k}$

Hence: . $\lim_{k\to\infty}a_n \;=\;\lim_{k\to\infty}\frac{1}{(1+\frac{1}{k})^k} \;=\;\frac{1}{\lim(1+\frac{1}{k})^k} \;=\;\frac{1}{e} \;\ne\;0$

Therefore, the series diverges.