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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

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

Well $\displaystyle {\lim _{k \to \infty }}{\left( {\frac{k}{{k + 1}}} \right)^k} = {\lim _{k \to \infty }}{\left( {1 + \frac{{ - 1}}{{k + 1}}} \right)^k}$

Re: Proving that a series diverges - help!

Alternatively, represent $\displaystyle \left(\frac{x}{x+1}\right)^x$ as $\displaystyle e^{x\ln(x/(x+1))}$ and then prove that $\displaystyle x\ln(x/(x+1))\to-1$ as $\displaystyle x\to\infty$ using either Taylor series or L'Hopital's rule.

Re: Proving that a series diverges - help!

Hello, kaya2345!

We are expected to know this definition:

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

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

Theorem: If the n^{th} term does *not* approach zero, the series diverges.

We have: .$\displaystyle 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: .$\displaystyle \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.