<s>[INST] Prove by induction that $\sum\limits_{i=1}^n \frac{1}{n+i} \leq \frac{3}{4}$ 
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 $\frac{3}{4}$. 
 [/INST] 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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