Convergent/Divergent, find sum

[iRaid]

Homework Statement

Determine whether the series is convergent or divergent. If it is convergent, find its sum.
$$\sum\limits_{n=1}^{\infty} (\frac{1}{e^n}+\frac{1}{n(n+1)})$$

Homework Equations

The Attempt at a Solution

So I found it's convergent:
$$\sum\limits_{n=1}^{\infty} ((\frac{1}{e})^{n}+\frac{1}{(n^{2}+n)})=\sum\limits_{n=1}^{\infty} ((\frac{1}{e})^{n}+\frac{1}{n}-\frac{1}{n+1})$$
$$\lim_{n \to \infty}((\frac{1}{e})^{n}+\frac{1}{n}-\frac{1}{n+1})=0+0+0$$
∴ the sum is convergent.

Now how would I find the sum of the series?

 

[Dick]

Homework Statement

Determine whether the series is convergent or divergent. If it is convergent, find its sum.
$$\sum\limits_{n=1}^{\infty} (\frac{1}{e^n}+\frac{1}{n(n+1)})$$

Homework Equations

The Attempt at a Solution

So I found it's convergent:
$$\sum\limits_{n=1}^{\infty} ((\frac{1}{e})^{n}+\frac{1}{(n^{2}+n)})=\sum\limits_{n=1}^{\infty} ((\frac{1}{e})^{n}+\frac{1}{n}-\frac{1}{n+1})$$
$$\lim_{n \to \infty}((\frac{1}{e})^{n}+\frac{1}{n}-\frac{1}{n+1})=0+0+0$$
∴ the sum is convergent.

Now how would I find the sum of the series?

Having a limit of 0 doesn't show it's convergent. (1/e)^n is geometric. The other two terms from a telescoping series. Find the sums separately.

 

[iRaid]

Having a limit of 0 doesn't show it's convergent. (1/e)^n is geometric. The other two terms from a telescoping series. Find the sums separately.

My professor said that a limit as n approaches infinity that equals zero means the series is convergent.. And I see what you're saying about the sum. Now I'm just confused on the convergent part.
Edit:
For the (1/e)^n I guess you could also say that since |1/e|<1, the series is convergent.

 

[Dick]

My professor said that a limit as n approaches infinity that equals zero means the series is convergent.. And I see what you're saying about the sum. Now I'm just confused on the convergent part.
Edit:
For the (1/e)^n I guess you could also say that since |1/e|<1, the series is convergent.

Right for the (1/e)^n. And I hope that's not really what your professor said. If the limit is NOT zero, then it's NOT convergent. Approaching zero doesn't mean it converges. 1/n approaches zero but the series 1/n is not convergent.

 

[iRaid]

Right for the (1/e)^n. And I hope that's not really what your professor said. If the limit is NOT zero, then it's NOT convergent. Approaching zero doesn't mean it converges. 1/n approaches zero but the series 1/n is not convergent.

Oh I see the problem here. 1/n is a harmonic series and divergent. Now how would I go about proving it is convergent?

 

[iRaid]

Also I just checked me notes it says the following:
If $\sum_{n=1}^{\infty} a_{n}$ is convergent, then $\lim_{n \to \infty}a_{n}=0$ and the opposite if a_n does not equal 0.

I think there is a difference in what I said and what my notes say right?

 

[Dick]

Oh I see the problem here. 1/n is a harmonic series and divergent. Now how would I go about proving it is convergent?

You know that the (1/e)^n part is convergent and you know it's sum. Now do the 1/n-1/(n+1) part. Like I said, it telescopes. If you know those two parts converge then the sum converges.

 

[Dick]

Also I just checked me notes it says the following:
If $\sum_{n=1}^{\infty} a_{n}$ is convergent, then $\lim_{n \to \infty}a_{n}=0$ and the opposite if a_n does not equal 0.

I think there is a difference in what I said and what my notes say right?

Yes, what you quoted doesn't say that if the limit is 0 then the sum converges.

 

[iRaid]

You know that the (1/e)^n part is convergent and you know it's sum. Now do the 1/n-1/(n+1) part. Like I said, it telescopes. If you know those two parts converge then the sum converges.

Ok then I did it correctly.
Thank you, again.

 

[iRaid]

Yes, what you quoted doesn't say that if the limit is 0 then the sum converges.

So you cannot say the opposite, if $\lim_{n \to \infty} a_{n}=0$ then $\sum_{n=1}^{\infty}a_{n}$ converges.?

 

[Dick]

So you cannot say the opposite, if $\lim_{n \to \infty} a_{n}=0$ then $\sum_{n=1}^{\infty}a_{n}$ converges.?

Doesn't the example of $a_n=1/n$ show that you can't?

 

[iRaid]

Doesn't the example of $a_n=1/n$ show that you can't?

I mean are there any other examples of this, other than the harmonic series?

 

[Dick]

I mean are there any other examples of this, other than the harmonic series?

Many. 1/n^(1/2) doesn't converge. 1/log(n) doesn't converge. log(n)/n doesn't converge. But the limits are all zero.

 

[iRaid]

Many. 1/n^(1/2) doesn't converge. 1/log(n) doesn't converge. log(n)/n doesn't converge. But the limits are all zero.

Alright, thanks a lot.