Converging Sums: Strategies for Solving Challenging Series

[twoflower]

Hi,

I have this sum:

<br /> \sum_{n = 1}^{\infty} \frac{n^2}{ \left( 2 + \frac{1}{n} \right)^{n}}<br />

I tried d'Alembert, I tried comparing it with \frac{1}{2^{n}}, but without success.

Could somebody point me to the right direction please?

Thank you.

 

[vincentchan]

\lim (2+1/n)^n = \lim (2(1+1/2(1/n))^n =\lim 2^n (1+1/n)^{n/2}
can you get from here?

 

[twoflower]

\lim (2+1/n)^n = \lim (2(1+1/2(1/n))^n =\lim 2^n (1+1/n)^{n/2}
can you get from here?

Thank you vincentchan, but I don't understand the second step - how can I get

<br /> \left(1+\frac{1}{n}\right)^{n/2}<br />

from

<br /> \left(1+\frac{1}{2}\left(\frac{1}{n}\right)\right)^{n}<br />

I know it must be some simple algebraic adjustment but I can't see that...

 

[vincentchan]

use change of variable, let u=n/2 the limit will become u->infinity/2, but infinity divided by 2 is also infinity... so u->infinity

 

[vincentchan]

OH... i made a mistake in post 2, the big idea is the same... see if you can catch it...

 

[Galileo]

You can also compare it with:

\sum_n \frac{n^2}{2^n}

 

[twoflower]

use change of variable, let u=n/2 the limit will become u->infinity/2, but infinity divided by 2 is also infinity... so u->infinity

I see it now (with being aware of the mistake you did :) ) I wouldn't see the possibility of the change at the first look, however.

 

[twoflower]

You can also compare it with:

\sum_n \frac{n^2}{2^n}

You're right, this is probably the easiest way. Thank you.