Why does $\sum_{n=1}^{\infty}{\frac{1}{n+1}}$ diverge?

Question

I guess it is kind of obvious that it diverges. But what divergent series do I compare it to?

score 4 · Answer 1 · answered Jul 17 '19 at 17:03

4

If you can use the fact that the harmonic series diverves, then, since$$\lim_{n\to\infty}\frac{\frac1{n+1}}{\frac1n}=1,$$your series diverges too.

Or you can use the fact that your series is the harmonic series minus its first term.

Another option here would be to use the integral test.

answered Jul 17 '19 at 17:03

"Or you can use the fact that your series is the harmonic series minus its first term."
How?
– Analysis Jul 17 '19 at 17:09
A series $a_0+a_1+a_2+a_3+\cdots$ converges if and only if the series $a_1+a_2+a_3+a_4+\cdots$ – José Carlos Santos Jul 17 '19 at 17:10

L F · Answer 2 · 2019-07-17T17:14:22.927

This is because the convergence of a series does not depend on one term, it means that you can rest one term and it will still converge

You have the divergent harmonic series without 1 term, lets see:

Suppose it converges to a real number $L$.

then $1+\displaystyle\sum_{n\ge1} \frac{1}{n+1}$ should converge to $L+1$

But $1+\displaystyle\sum_{n\ge1} \frac{1}{n+1}=\displaystyle\sum_{n\ge1} \dfrac{1}{n}$ and it should be equal to $L+1$, which is obviously false

score 2 · Answer 3 · answered Jul 17 '19 at 17:14

2

Compare to the harmonic series. Note that $$\frac{1}{n+1}=\frac 1n-\frac{1}{n(n+1)}.$$

answered Jul 17 '19 at 17:14

Allawonder

3 Answers3