# Thread: converge or diverge problem

1. ## converge or diverge problem

Hi! Can someone help me figure out these problems please? I have to determine whether the series converge or diverge and give a reason to the answer. Thanks!

1.)

Σ [ln (n+1)]/(n+1)
n=2

2.)

Σ 2/ (1+e^n)
n=1

2. Originally Posted by cherry106
Hi! Can someone help me figure out these problems please? I have to determine whether the series converge or diverge and give a reason to the answer. Thanks!

1.)

Σ [ln (n+1)]/(n+1)
n=2
Diverges as $\frac{\ln(n+1)}{n+1}>\frac{1}{n+1}$ for $n>e-1$ and $\sum_{n=2}^\infty \frac{1}{n+1}$ diverges
as it is just the tail of the harmonic series.

RonL

3. Originally Posted by cherry106
2.)

Σ 2/ (1+e^n)
n=1
Converges, as $1+e^n>n^2$ for $n>0$, so $\frac{1}{1+e^n}<\frac{1}{n^2}$ and $\sum_{n=1}^{\infty } \frac{1}{n^2}$ converges.

RonL

4. Thanks for the help!

For problems like that how do you know what to compare the equation to? for example in question one you compared the denominator to {1}/{n+1} for n>e-1 and for question two you used n^2 for n>0.

5. Originally Posted by cherry106
Thanks for the help!

For problems like that how do you know what to compare the equation to? for example in question one you compared the denominator to {1}/{n+1} for n>e-1 and for question two you used n^2 for n>0.
In question one we have something slowly increasing over n+1, so the whole
thing decreases more slowly that 1/(n+1) so we try comparing the series with
the harmonic series.

In the second we have essentially 1/(1+e^n), but e^n increases more
rapidly than any power of n, so 1/(1+e^n) decreases more rapidly than
the reciprocal of any power of n, and as

$\sum 1/n^2$

is the first of these series which converges its what we use for a comparison
series.

RonL