Finding the maximum value of a divergent series

Question

I came across this divergent sum-

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

Now,a divergent sum does not a limit.So is it possible to get a maximum value for the sum or more specifically prove that the series is lesser than a particular value?

More specifically I got this idea from this answer-How do I prove $\frac 34\geq \frac{1}{n+1}+\frac {1}{n+2}+\frac{1}{n+3}+\cdots+\frac{1}{n+n}$

Where am I getting my concept wrong?

Thanks for any help!!

Answer 1

By approximating the partial sums by definite integrals (see here), we have that $$ \log(N+2)-\log2=\int_1^{N+1}\frac1{x+1}\mathrm dx\le\sum_{n=1}^N\frac1{n+1}\le\int_0^N\frac1{x+1}\mathrm dx=\log(N+1) $$ for $N\ge1$. Hence, we can bound the partial sums from above with $\log(N+1)$. However, this bound depends on the number of terms in the partial sums and goes to $\infty$ as $N\to\infty$. Also, we see that $$ \lim_{N\to\infty}\biggl[\frac1{\log N}\sum_{n=1}^N\frac1{n+1}\biggr]=1. $$ I hope this helps.