1. [SOLVED] 2 infinite series

I wasn't able to find if they converges or not :
1) $\sum_{n=1}^{+\infty} \left( \frac{n}{2n+1}\right) ^n$ My intuition tells me it converges.
And 2) $\sum_{n=1}^{+\infty} (\sqrt[n]{n}-1)^n$. This one is going to make me crazy. I'm not sure at all if it diverges or not.
For both I've tried the root test, integral test and for the first the ratio test but in each case I failed to get a conclusion. I need a little help (maybe the test to do, and any trick I might not see) but not the full detailed answer, please.

2. Originally Posted by arbolis
I wasn't able to find if they converges or not :
1) $\sum_{n=1}^{+\infty} \left( \frac{n}{2n+1}\right) ^n$ My intuition tells me it converges.
And 2) $\sum_{n=1}^{+\infty} (\sqrt[n]{n}-1)^n$. This one is going to make me crazy. I'm not sure at all if it diverges or not.
For both I've tried the root test, integral test and for the first the ratio test but in each case I failed to get a conclusion. I need a little help (maybe the test to do, and any trick I might not see) but not the full detailed answer, please.
1) Use the Root test $
\sqrt[n]{{\left( {\tfrac{n}
{{2n + 1}}} \right)^n }} = \tfrac{n}
{{2n + 1}} \to \tfrac{1}
{2}
$
so it converges

2) Again, the root test. Note that: $
\sqrt[n]{n} - 1 = e^{\tfrac{{\ln \left( n \right)}}
{n}} - 1 = \tfrac{{\ln \left( n \right)}}
{n} + \tfrac{{\ln ^2 \left( n \right)}}
{{2 \cdot n^2 }} + \tfrac{{\ln ^3 \left( n \right)}}
{{3! \cdot n^3 }} + ...
$

3. Sorry for the first one! I realize it was much easier than I thought.
2) Again, the root test. Note that:
Thanks! I almost got something similar, but couldn't find it. Correct me if I'm wrong, but $\frac{\ln (n)}{n} \rightarrow 0$, so $e^{\frac{\ln (n)}{n}} \rightarrow 1$, therefore $e^{\frac{\ln (n)}{n}}-1 \rightarrow 0$, so the series converges by the root test.

4. Originally Posted by arbolis
Sorry for the first one! I realize it was much easier than I thought.
Thanks! I almost got something similar, but couldn't find it. Correct me if I'm wrong, but $\frac{\ln (n)}{n} \rightarrow 0$, so $e^{\frac{\ln (n)}{n}} \rightarrow 1$, therefore $e^{\frac{\ln (n)}{n}}-1 \rightarrow 0$, so the series converges by the root test.
That's right

5. Originally Posted by arbolis
I wasn't able to find if they converges or not :
1) $\sum_{n=1}^{+\infty} \left( \frac{n}{2n+1}\right) ^n$ My intuition tells me it converges.
And 2) $\sum_{n=1}^{+\infty} (\sqrt[n]{n}-1)^n$. This one is going to make me crazy. I'm not sure at all if it diverges or not.
For both I've tried the root test, integral test and for the first the ratio test but in each case I failed to get a conclusion. I need a little help (maybe the test to do, and any trick I might not see) but not the full detailed answer, please.
You tell me the answer, but an even easier way of doing the first one is that,

$\left(\frac{n}{2n+1}\right)^n\sim~?$

6. You tell me the answer, but an even easier way of doing the first one is that,

$\left( \frac{1}{2}\right) ^n$, or I don't know. Whatever, I prefer to be rigorous, otherwise I won't do well in the final exam .

7. Originally Posted by Mathstud28
You tell me the answer, but an even easier way of doing the first one is that,

$\left(\frac{n}{2n+1}\right)^n\sim~?$
$
\left( {\tfrac{n}
{{2n + 1}}} \right)^n = \left( {\tfrac{{\tfrac{1}
{2}}}
{{1 + \tfrac{1}
{{2n}}}}} \right)^n = \tfrac{{\left( {\tfrac{1}
{2}} \right)^n }}
{{\left( {1 + \tfrac{1}
{{2n}}} \right)^n }}
$

Now $
{\left( {1 + \tfrac{1}
{{2n}}} \right)^n } \to {\sqrt{e}}
$
as n tends to infinity (remember that: $
\left( {1 + \tfrac{x}
{n}} \right)^n \to e^x
$
)

So $
\left( {\tfrac{n}
{{2n + 1}}} \right)^n \sim \tfrac{1}
{{\sqrt e }} \cdot \left( {\tfrac{1}
{2}} \right)^n
$

8. Originally Posted by arbolis
$\left( \frac{1}{2}\right) ^n$, or I don't know. Whatever, I prefer to be rigorous, otherwise I won't do well in the final exam .
No thats totally fine, I was just saying, you can show also pretty easily (and a little more pecularily, which is what I like) that

$\lim_{x\to\infty}\frac{\left(\frac{n}{2n+1}\right) ^n}{\left(\frac{1}{2}\right)^n}=e^{\frac{-1}{2}}\Rightarrow\left(\frac{n}{2n+1}\right)^n\sim {e^{\frac{-1}{2}}\left(\frac{1}{2}\right)^n}$

Which is obviously convergent.

9. Originally Posted by PaulRS
$
\left( {\tfrac{n}
{{2n + 1}}} \right)^n = \left( {\tfrac{{\tfrac{1}
{2}}}
{{1 + \tfrac{1}
{{2n}}}}} \right)^n = \tfrac{{\left( {\tfrac{1}
{2}} \right)^n }}
{{\left( {1 + \tfrac{1}
{{2n}}} \right)^n }}
$

Now $
{\left( {1 + \tfrac{1}
{{2n}}} \right)^n } \to {\sqrt{e}}
$
as n tends to infinity (remember that: $
\left( {1 + \tfrac{x}
{n}} \right)^n \to e^x
$
)

So $
\left( {\tfrac{n}
{{2n + 1}}} \right)^n \sim \tfrac{1}
{{\sqrt e }} \cdot \left( {\tfrac{1}
{2}} \right)^n
$
Good timing