1. ## series convergence

Series a_n converges and a_n>=0. Does series e^{a_n}-1 converge as well?

I suppose sum {e^{a_n}} < e^{sum{a_n}} => the sequence of partial sums converges and the series should converge. But lim e^{a_n} /= 0!What is wrong ?

The same question about convergence for series n*a_{n^2}.

Thanks for any help=)

2. Originally Posted by dim
Series a_n converges and a_n>=0. Does series e^{a_n}-1 converge as well?

I suppose that sum(e^{a_n}) < e^(sum{a_n}) => the sequence of partial sums converges and the series should converge. But lim e^{a_n} /=0! What is wrong ?

The same question about convergence for series n*a_{n^2}.

Thanks for any help=)
$\displaystyle e^{a_n}-1=1+a_n+\frac{a_n^2}{2}+\cdots-1=a_n+\frac{a_n^2}{2}+\cdots=\sum_{N=1}^{\infty}\f rac{a_n^N}{N!}$

Can you go from there?

3. Indeed I cannot use series expansion yet! Is there any other way?

4. Originally Posted by dim
Indeed I cannot use series expansion yet! Is there any other way?
Yeah I thought of another way. Consider $\displaystyle \lim_{n\to\infty}\frac{e^{a_n}-1}{a_n}$. Now let $\displaystyle \varphi=a_n$ so because the series $\displaystyle \sum{a_n}$ converges we know that $\displaystyle a_n$ is a null sequence (it converges to zero). So we see that as $\displaystyle n\to\infty\implies{\varphi\to{0}}$. So our limit becomes $\displaystyle \lim_{\varphi\to{0}}\frac{e^{\varphi}-1}{\varphi}=1$. Thus the two series $\displaystyle \sum\left\{{e^{a_n}-1}\right\}$ and $\displaystyle \sum{a_n}$ share convergence.

5. Originally Posted by dim
Series a_n converges and a_n>=0.
The same question about convergence for series n*a_{n^2}.

Thanks for any help=)
Is there any more info about the second one? If you have some extra restrictions on the set you should say. I have come up with a solution, but it only applies if you have a few more restrictions, report back if this is the exact question.

Consider $\displaystyle \sum{na_{n^2}}$, we know that $\displaystyle a_n=0$, now ASSUMING that $\displaystyle a_n\in\downarrow$ and $\displaystyle a_n\in\mathcal{C}$ we have that this series and the integral $\displaystyle \int_{0}^{\infty}na_{n^2}dn$ share convergence. Now let $\displaystyle \psi=n^2$ then we have that

$\displaystyle \frac{1}{2}\int_0^{\infty}a_{\psi}d\psi$, now since we are given that $\displaystyle \sum{a_n}$ converges so does $\displaystyle \frac{1}{2}\int_0^{\infty}a_{\psi}d\psi\quad\there fore\sum{na_{n^2}}\text{ converges}\quad\blacksquare$

6. Originally Posted by Mathstud28
No other restrictions, only that a_n is positive and converges (series).
I thought about finding $\displaystyle \lim_{n\to\infty}\frac{(n+1)a_{(n+1)^2}}{na_{n^2}}$. We know that $\displaystyle \lim_{n\to\infty}\frac{a_{n^2+1}}{a_{n^2}}=0$ as series a_n converges. Can we say that $\displaystyle \lim_{n\to\infty}\frac{a_{(n+1)^2}}{a_{n^2}}=0$? If so we could probably use this test.

P.S. And I see we can by applying chain multiplication $\displaystyle \lim_{n\to\infty}\frac{a_{(n+1)^2}}{a_{n^2}}=\lim_ {n\to\infty}\frac{a_{n^2+1+2n}}{a_{n^2+1+2n-1}}*\frac{a_{n^2+1+2n-1}}{a_{n^2+1+2n-2}}*...*\frac{a_{n^2+1}}{a_{n^2}}=0$

7. Originally Posted by Mathstud28
Yeah I thought of another way. Consider $\displaystyle \lim_{n\to\infty}\frac{e^{a_n}-1}{a_n}$. Now let $\displaystyle \varphi=a_n$ so because the series $\displaystyle \sum{a_n}$ converges we know that $\displaystyle a_n$ is a null sequence (it converges to zero). So we see that as $\displaystyle n\to\infty\implies{\varphi\to{0}}$. So our limit becomes $\displaystyle \lim_{\varphi\to{0}}\frac{e^{\varphi}-1}{\varphi}=1$. Thus the two series $\displaystyle \sum\left\{{e^{a_n}-1}\right\}$ and $\displaystyle \sum{a_n}$ share convergence.
Thanks a lot! It is great!

There is one more question I could not deal with .. I need to show that $\displaystyle \sum{\frac{1}{nlnn}}$ is convergent, but I can not use integral test for convergence! If you have any idea, I will highly appreciate it.

8. Originally Posted by dim
No other restrictions, only that a_n is positive and converges (series).
I thought about finding $\displaystyle \lim_{n\to\infty}\frac{(n+1)a_{(n+1)^2}}{na_{n^2}}$. We know that $\displaystyle \lim_{n\to\infty}\frac{a_{n^2+1}}{a_{n^2}}=0$ as series a_n converges. Can we say that $\displaystyle \lim_{n\to\infty}\frac{a_{(n+1)^2}}{a_{n^2}}=0$? If so we could probably use this test.

P.S. And I see we can by applying chain multiplication $\displaystyle \lim_{n\to\infty}\frac{a_{(n+1)^2}}{a_{n^2}}=\lim_ {n\to\infty}\frac{a_{n^2+1+2n}}{a_{n^2+1+2n-1}}*\frac{a_{n^2+1+2n-1}}{a_{n^2+1+2n-2}}*...*\frac{a_{n^2+1}}{a_{n^2}}=0$
I do not have time now to do check this out thourougly but consider that $\displaystyle \lim_{n\to\infty}\frac{a_{(n+1)^2}}{a_{n^2}}$ is the Ratio test for this series.
Originally Posted by dim
Thanks a lot! It is great!

There is one more question I could not deal with .. I need to show that $\displaystyle \sum{\frac{1}{nlnn}}$ is convergent, but I can not use integral test for convergence! If you have any idea, I will highly appreciate it.
Use Cauchy's Condensation test. A series satisfying the same restrictions (save continuity) of the integral test converges iff $\displaystyle \sum{2^na_{2^n}}$ converges, so now consider $\displaystyle \frac{2^n}{2^n\ln\left(2^n\right)}$. Use your laws of logarithims and this becomes the Harmonic Series, thus both this and your question series diverge by CC's test.

9. A series does not converge unless the sequence of numbers converges to 0. In particular, $\displaystyle \sum e^{a_n}- 1$ cannot converge if $\displaystyle e^{a_n}$ does not converge to 1 which is the same as saying that $\displaystyle a_n$ converges to 0.

Of course, if $\displaystyle a_n$ does converge to 0, it does not follow that [tex]\sum e^{a_n}-1[tex] converges.

10. Originally Posted by HallsofIvy
A series does not converge unless the sequence of numbers converges to 0. In particular, $\displaystyle \sum e^{a_n}- 1$ cannot converge if $\displaystyle e^{a_n}$ does not converge to 1 which is the same as saying that $\displaystyle a_n$ converges to 0.

Of course, if $\displaystyle a_n$ does converge to 0, it does not follow that [tex]\sum e^{a_n}-1[tex] converges.
What does that have to do with anything?