Does $\sum_{n=1}^{\infty} \frac{1}{n+1}$ converge?

Question

Sorry for the oversimplified question, but does the series $\sum_{n=1}^{\infty} \frac{1}{n+1}$ converge? The ratio test of it gives the result of "1". Thanks a lot.

Answer 1

If the ratio test gives $1$ then we are in a situation where we can't conclude wether the series is convergent or divergent.

There are several ways to prove that your given series is divergent and one way is: let the partial sum be

$$S_n=\sum_{k=1}^n \frac{1}{1+k}$$

Then we have

$$S_{2n}-S_n=\sum_{k=n+1}^{2n}\frac{1}{1+k}\geq n\times\frac{1}{1+2n}\geq\frac{1}{3}\tag{1}$$

If the series were convergent, then the sequence $(S_{2n}-S_n)$ would have been convergent to $0$. This is not the case, as shown by the inequality $(1)$.

Answer 2

You can use the comparison test:

$$\frac{1}{2}\cdot \sum_{n=1}^\infty \frac{1}{n}\leqslant \sum_{n=1}^\infty \frac{1}{n+1} $$

And by the divergence of the harmonic sum we conclude that our sum diverges.

Answer 3

This is not a convergent sequence so it is not possible to find the sum.

Let $\displaystyle S=\sum_{i=1}^{\infty}\frac{1}{i}=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\dots= (1)+(\frac{1}{2}+\frac{1}{3})+(\frac{1}{4}+\frac{1}{5}+\dots+\frac{1}{7})+\dots+(\frac{1}{2^k}+\frac{1}{2^k+1}+\dots+\frac{1}{2^{k+1}-1})+\dots\ge 1+(\frac{1}{4}+\frac{1}{4})+(\frac{1}{8}+\frac{1}{8}+\dots+\frac{1}{8})+\dots+(\frac{1}{2^{k+1}}+\frac{1}{2^{k+1}}+\dots+\frac{1}{2^{k+1}})+\dots=1+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\dots+2^k\frac{1}{2^{k+1}}+\dots >1+\frac{n}{2}$

(for any $n\in N$)

So $S$ cannot converge.

Now $\displaystyle S=\sum_{i=1}^{\infty}\frac{1}{i}=1+\sum_{i=1}^{\infty}\frac{1}{(i+1)}$

As $S$ is not convergent so $\displaystyle\sum_{i=1}^{\infty}\frac{1}{(i+1)}$ is also not convergent(rather they both diverge).