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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!!

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Soham
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  • Your question is questionable. If there were a maximum value the series would be convergent by the monotonic convergence theorem. – Lutz Lehmann Feb 09 '16 at 09:28
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    If you can prove that the sum is smaller than a particular value, then it will converge because the sequence of partial sums is monotonically increasing. The question that you mentioned involves a finite sum. – Tunococ Feb 09 '16 at 09:29
  • @LutzL But this series is much like the (harmonic series)-1....and harmonic series in divergent.... – Soham Feb 09 '16 at 09:31
  • Yes, that's what You wrote. The finite sum converges to $\ln 2$ and the series can be divided into an infinity of those finite sums. – Lutz Lehmann Feb 09 '16 at 09:36
  • @tatan : Your concept is wrong because you compare an infinite series to a finite series. Of course, in case of the finite series, which is increassing as long as you add terms, when you stop to add, the series reaches a maximum value. In the case of the infinite series you never stop to add more terms and there is no maximum value if the series is not convergent (your example), or tends to a limit value if the series is convergent (not your example, but with other examples). – JJacquelin Feb 09 '16 at 09:40

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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.

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