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Prove by induction that $\sum\limits_{i=1}^n \frac{1}{n+i} \leq \frac{3}{4}$.

I have to prove this inequality using induction, I proved it for $n=1$ and now I have to prove it for $n+1$ assuming $n$ as hypothesis, but this seems impossible to me because the difference between the sum of $n$ and the sum of $n+1$ is a positive value. Adding a positive value on both sides of the inequality, I don't know how to prove that is always less than or equal to $\frac{3}{4}$.

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    $\begingroup$ Let $S_n=\sum\limits_{i=1}^n \frac{1}{n+i}$ then $S_{n+1}=S_n+\frac1{2n+2}+\frac1{2n+1}-\frac1{n+1}$ hence $S_{n+1}+\frac1{2n+4}=S_n+\frac1{2n+2}-R_n$ with $R_n=\frac1{n+1}-\frac1{2n+4}-\frac1{2n+1}=\frac{n-1}{(2n+4)(2n+1)(n+1)}\leqslant0$. Thus, for every $n\geqslant1$, $S_n+\frac1{2n+2}\leqslant S_1+\frac14=\frac34$, which implies that $S_n\leqslant\frac34-\frac1{2n+2}<\frac34$, qed. This is a direct proof. To prove the desired result by recurrence from scratch, prove instead by recurrence the slight improvement $S_n\leqslant\frac34-\frac1{2n+2}$ for every $n\geqslant1$. $\endgroup$ – Did Oct 18 '16 at 14:57
  • $\begingroup$ @Did, this comment should be an answer... $\endgroup$ – lhf Oct 18 '16 at 14:58
  • $\begingroup$ @Did, I think the same as lhf: your answer is closer to the original question! $\endgroup$ – Darío G Oct 18 '16 at 15:09
  • $\begingroup$ In fact, the sum is less than $\frac{25}{36}$. See here for an elementary proof. $\endgroup$ – TonyK Oct 18 '16 at 16:32
  • $\begingroup$ The trick is realizing that you will have to prove something slightly stronger. If $\sum^n a_i \le 3/4$ then $\sum^{n+1} a_i = \sum^n a_i + a_{n+1} \le 3/4 + a_{n+1}$ and as $a_{n+1} > 0$ you will never be able to prove the desired result. But if you attempt to prove $\sum^n a_i \le 3/4 - b_n$ the $\sum^{n+1} a_i \le 3/4 - b_n + a_{n+1}$ and it's a matter of showing $a_{n+1} - b_n \ge -b_{n+1}$. $\endgroup$ – fleablood Oct 18 '16 at 17:12
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Define $\displaystyle{S_n=\sum_{i=1}^n \dfrac{1}{n+i}}$. Then we have for every $n$ that:

\begin{align*} S_{n+1}-S_n&=\sum_{i=1}^{n+1}\dfrac{1}{n+1+i}-\sum_{i=1}^n \dfrac{1}{n+i}\\ &=\sum_{i=1}^{n+1}\dfrac{1}{n+1+i} - \left(\sum_{i=2}^n \dfrac{1}{n+i}\right)-\dfrac{1}{n+1}\\ &=\sum_{i=1}^{n+1}\dfrac{1}{n+1+i} - \left(\sum_{j=1}^{n-1} \dfrac{1}{n+1+j}\right)-\dfrac{1}{n+1}\\ &=\dfrac{1}{n+1+(n+1)}+\dfrac{1}{n+1+n}-\dfrac{1}{n+1}\\ &=\dfrac{1}{2n+1}-\dfrac{1}{2(n+1)}=\dfrac{1}{2(n+1)(2n+1)} \end{align*}

In particular, the sequence $S_n$ can also be defined by recurrence with the formula $$\begin{cases}S_1=\dfrac{1}{2}\\ S_{n+1}=S_n+\dfrac{1}{2(n+1)(2n+1)}\end{cases}$$

and so, $S_n$ is simply defined by the formula $$S_n=\sum_{i=1}^n \dfrac{1}{2n(2n-1)}.$$

Finally, showing that $S_n\leq \frac{3}{4}$ for all $n$ is the same as proving that the series $\displaystyle{\sum_{n=1}^\infty \dfrac{1}{2n(2n-1)}}$ converges to a number less than or equal to $\dfrac{3}{4}$.

If we focus on the infinite series, notice that $\dfrac{1}{2n(2n-1)}=\dfrac{1}{2n-1}-\dfrac{1}{2n}$, and so we have:

$$\sum_{n=1}^\infty \dfrac{1}{2n(2n-1)}=\sum_{n=1}^\infty \left(\dfrac{1}{2n-1}-\dfrac{1}{2n} \right)= \sum_{k=1}^\infty \dfrac{(-1)^{k+1}}{k}$$

I know the last series converges (by the test of alternating series), but I am just not sure about the value.

EDIT: AS lhf points out in his comment, the value of the last sum is $\log 2=0.69314\ldots$, which is surely less than $\dfrac{3}{4}$.

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  • $\begingroup$ The value of the last sum is $\log 2$. $\endgroup$ – lhf Oct 18 '16 at 15:00
  • $\begingroup$ @lhf Thanks. I will add it to my answer if you don't mind. $\endgroup$ – Darío G Oct 18 '16 at 15:01
  • $\begingroup$ @W Nop: $\;\log2+0.3010=0.9... >\frac34\;$ Now, we have that $\;\log 2=0.6...<\frac34\;$ , that's true. $\endgroup$ – DonAntonio Oct 18 '16 at 15:18
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    $\begingroup$ @DonAntonio Thanks for noticing the typo. + was actually and equal. $\endgroup$ – Darío G Oct 18 '16 at 15:19
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Preliminary Result

Note that $$ \begin{align} s_{n+1}-s_n &=\sum_{k=1}^{n+1}\frac1{n+k+1}-\sum_{k=1}^n\frac1{n+k}\\ &=\sum_{k=2}^{n+2}\frac1{n+k}-\sum_{k=1}^n\frac1{n+k}\\ &=\frac1{2n+2}+\frac1{2n+1}-\frac1{n+1}\\ &=\frac1{2n+1}-\frac1{2n+2}\tag{1} \end{align} $$ Therefore, since $\frac1{2k}-\frac1{2k+1}\ge\frac12\left(\frac1{2k}-\frac1{2k+2}\right)$ $$ \begin{align} s_n &=\sum_{k=1}^n\left(\frac1{2k-1}-\frac1{2k}\right)\\ &=1-\sum_{k=1}^{n-1}\left(\frac1{2k}-\frac1{2k+1}\right)-\frac1{2n}\\ &\le1-\frac12\sum_{k=1}^{n-1}\left(\frac1{2k}-\frac1{2k+2}\right)-\frac1{2n}\\[3pt] &=\frac34-\frac1{4n}\tag{2} \end{align} $$


Inductive Proof

We will prove inductively that $$ \sum_{k=1}^n\frac1{n+k}\le\frac34-\frac1{4n}\tag{3} $$ Note that $(3)$ holds, with equality, for $n=1$.

Suppose that $$ \sum_{k=1}^n\frac1{n+k}\le\frac34-\frac1{4n}\tag{4} $$ Note that $\frac1{4n+4}\le\frac1{4n}-\frac1{2n+1}+\frac1{2n+2}\iff\frac1{2n+1}\le\frac12\left(\frac1{2n}+\frac1{2n+2}\right)$, which is true because $\frac1x$ is convex. Therefore, $$ \begin{align} \sum_{k=1}^{n+1}\frac1{n+k+1} &=\sum_{k=2}^{n+2}\frac1{n+k}\\ &=\color{#00A000}{\sum_{k=1}^n\frac1{n+k}}\color{#C00000}{-\frac1{n+1}+\frac1{2n+1}+\frac1{2n+2}}\\ &\le\color{#00A000}{\frac34-\frac1{4n}}\color{#C00000}{+\frac1{2n+1}-\frac1{2n+2}}\\[6pt] &\le\frac34-\frac1{4n+4}\tag{5} \end{align} $$ which completes the induction.

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Assuming $n>1$, $$\begin{align} \sum_{i=1}^n \frac 1{n+i} =\frac 1n \sum_{i=1}^n \frac 1{1+\frac in} &\color{red}<\int_0^1 \frac 1{1+x} \;\; dx = \bigg[\ln (1+x)\bigg]_0^1 =\ln 2=0.693 \color{red}<\frac 34\\ \Rightarrow \sum_{i=1}^n \frac 1{n+i}&\color{red}<\frac 34\qquad\blacksquare\end{align}$$

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  • $\begingroup$ You carried the $\frac1n$ (representing $\mathrm{d}x$) into the integral (after the first $\color{#C00000}{\lt}$). Thus, you should not have $\frac1n\log(2)$, but simply $\log(2)$. $\endgroup$ – robjohn Oct 18 '16 at 17:07
  • $\begingroup$ @robjohn - Yes, that's correct. Thanks for pointing it out. Edited accordingly. $\endgroup$ – hypergeometric Oct 18 '16 at 17:11
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Note that you do not use induction in the above. Another way: Define $$S_n := \sum_{i=1}^n \frac{1}{n+i}$$ Then $$S_{n+1} = \sum_{i=1}^{n+1} \frac{1}{n+1+i} = \sum_{i=1}^{n+1} \frac{1}{n+(i+1)} = \sum_{i=2}^{n+2} \frac{1}{n+i}.$$ Hence $$S_{n+1} = S_n - \frac{1}{n+1} + \frac{1}{n+n+1} +\frac{1}{n+n+2}.$$ Now proceed...

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Let $$S_n=\sum_{i=1}^{n}\frac{1}{n+i}=\frac{1}{n}\sum_{k=1}^{n}\frac{1}{1+\frac{k}{n}}=H_{2n}-H_n.\tag{1}$$ We have $$ S_{n+1}-S_n = \frac{1}{2n+2}+\frac{1}{2n+1}-\frac{1}{n+1} = \frac{1}{(2n+1)(2n+2)} \tag{2}$$ hence $\{S_n\}_{n\geq 1}$ is an increasing sequence. Due to $(1)$ and Riemann sums, $$S_n \leq \lim_{n\to +\infty}S_n = \int_{0}^{1}\frac{dx}{1+x} = \log(2). \tag{3}$$ Over the interval $(0,1)$, the function $f(x)=x(1-x)$ is positive and $\leq \frac{1}{4}$, hence $$ \frac{1}{16}\geq \int_{0}^{1}\frac{x^2(1-x)^2}{1+x}\,dx = -\frac{11}{4}+4\log 2\tag{4} $$ and $$\log(2)\leq \frac{45}{64}<\frac{3}{4}.\tag{5}$$

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  • $\begingroup$ What's the downvote for? I won't ever understand this kind of behaviour. If you do not like to learn some new technique or math, just ignore it. What is the point in downvoting? That goes against increasing the community knowledge. $\endgroup$ – Jack D'Aurizio Oct 18 '16 at 18:42
  • $\begingroup$ Well, it wasn't me, but using logs for this question is silly if you ask me. $\endgroup$ – TonyK Oct 18 '16 at 23:10
  • $\begingroup$ @TonyK: it is not my fault that $$ H_n = \sum_{k=1}^{n}\frac{1}{k}\sim \log(n),$$ and that $\lim_{n\to +\infty}S_n=\log(2)$, if you ask me. $\endgroup$ – Jack D'Aurizio Oct 18 '16 at 23:13

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