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Let $n\in \mathbb N^{+}$,show that $$\dfrac{1}{2+1}+\dfrac{2}{2^2+2}+\dfrac{3}{2^3+3}+\cdots+\dfrac{n}{2^n+n}<\dfrac{3}{2}$$

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    $\begingroup$ Please show us what you have tried/your own thoughts. $\endgroup$
    – mickep
    Oct 8 '15 at 11:39
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Using that $\sum_{n=k}^\infty n 2^{-n} = (k+1)2^{1-k}$, $$ \sum_{n=1}^\infty \frac{n}{2^n + n} < \frac{1}{3} + \frac{2}{6} + \frac{3}{11} + \frac{4}{20} + \frac{5}{37} + \sum_{n=6}^\infty \frac{n}{2^n}\\ = \frac{1}{3} + \frac{2}{6} + \frac{3}{11} + \frac{4}{20} + \frac{5}{37} + \frac{7}{32} = \frac{291727}{195360} < \frac{3}{2}. $$

Also ugly, but also works.

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  • $\begingroup$ Not ugly at all! +1 $\endgroup$
    – yurnero
    Oct 8 '15 at 11:41
  • $\begingroup$ @zhoraster I don't see how you can justify the proof of the last step without at least some cumbersome computation. You've left out some crucial steps at the end. That's not good enough! $\endgroup$
    – user387350
    Dec 20 '16 at 17:08
  • $\begingroup$ @RumpyPumpy, shame on me and on those 10 persons who had upvoted! $291727\cdot 2 = 583... < 586.. =195360\cdot 3$. $\endgroup$
    – zhoraster
    Dec 20 '16 at 17:42
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I will suggest an ugly solution and I hope someone will refine it.

The ugly part is that you should check your inequality for $1\leq n\leq 8$.

After doing that you can prove using $2$ induction that for $n\geq 9$ the following stronger inequality holds:

$$\sum \frac{n}{2^n+n}\leq \frac{3}{2}-\frac{1}{n}.$$

I admit, it's ugly but it works.

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$$\sum_{n\geq 1}\frac{n}{2^n+n} = \frac{31}{33}+\sum_{n\geq 4}\sum_{k\geq 1}(-1)^{k+1} n^k 2^{-nk}\leq \frac{31}{33}+\frac{5}{8}-\frac{173}{1728}+\frac{25157}{1229312}<\frac{3}{2}. $$

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