0
$\begingroup$

For all integers $n\geq2$,

$$\sum_{k=1}^n\frac{1}{n+k} \geq \frac{7}{12}$$

How would you prove this by induction?

  • 0
    $\frac 1 {n+1+k} < \frac 1 {n+k} $. Do you know by how much. Can you prove $\sum (\frac 1 {n+1+k} - \frac 1 {n+k} + \frac 1 {2 (n+1)} \le 0$? – 2017-01-26

3 Answers 3

3

$$\sum_{k=1}^{n}\frac{1}{n+k} = H_{2n}-H_{n}$$ are the terms of an increasing sequence, since $$ H_{2n+2}-H_{n+1}-H_{2n}+H_n = \frac{1}{2n+2}+\frac{1}{2n+1}-\frac{1}{n+1} = \frac{1}{(2n+1)(2n+2)}>0.$$

  • 1
    I don't see how this helps though? – 2017-01-26
  • 0
    @Ivy: $\sum_{k=1}^{2}\frac{1}{2+k}=\frac{7}{12}$ and I have just proved the next terms are bigger. – 2017-01-26
0

By induction you can do:

$$\sum_{k=1}^{n+1}\frac{1}{n+1+k}=\sum_{k=1}^n\frac{1}{n+k} -\frac{1}{n+1}+\frac{1}{2n+1}+\frac{1}{2n+2}=\\ =\sum_{k=1}^n\frac{1}{n+k}+\frac{1}{(2n+1)(2n+2)}\ge \frac{7}{12}+\frac{1}{(2n+1)(2n+2)}\ge \frac{7}{12}$$

0

If $s_n = \sum_{k=1}^n\frac{1}{n+k} $, then

$\begin{array}\\ s_{n+1} &= \sum_{k=1}^{n+1}\dfrac{1}{n+1+k}\\ &= \sum_{k=2}^{n+2}\dfrac{1}{n+k}\\ &= \sum_{k=2}^{n}\dfrac{1}{n+k}+\dfrac1{2n+1}+\dfrac1{2n+2}\\ &= \sum_{k=1}^{n}\dfrac{1}{n+k}-\dfrac1{n+1}+\dfrac1{2n+1}+\dfrac1{2n+2}\\ &= s_n+\dfrac{(2n+1)+(2n+2)-2(2n+1)}{(2n+1)(2n+2)}\\ &= s_n+\dfrac{1}{(2n+1)(2n+2)}\\ \end{array} $

Therefore $s_{n+1} > s_n$, so $s_n$ is increasing.

Therefore, if $n > m$, $s_n > s_m$.

Note that we can bound $s_n$ because

$\begin{array}\\ s_{n+1} &= s_n+\dfrac{1}{(2n+1)(2n+2)}\\ &< s_n+\dfrac{1}{(2n)(2n+2)}\\ &< s_n+\frac14\dfrac{1}{n(n+1)}\\ &< s_n+\frac14(\dfrac1{n}-\dfrac1{n+1})\\ \end{array} $

so that $s_{n+1}-s_n < \frac14(\dfrac1{n}-\dfrac1{n+1}) $.

Summing, $\sum_{k=1}^{m-1}(s_{n+k+1}-s_{n+k}) < \sum_{k=1}^{m-1}\frac14(\dfrac1{n}-\dfrac1{n+1}) $ or $s_{n+m}-s_n \lt \frac14(\frac1{n}-\frac1{n+m}) \lt \frac1{4n} $.

Therefore, $\lim_{n \to \infty} s_n$ exists, and, if $S=\lim_{n \to \infty} s_n$, $S < s_n +\dfrac1{4n}$.

We can similarly get a lower bound on $s_n$ and, therefore, a lower bound on $S$ in terms of $n$ and $s_n$.

(the following is done with copy, paste, and edit.)

$\begin{array}\\ s_{n+1} &= s_n+\dfrac{1}{(2n+1)(2n+2)}\\ &> s_n+\dfrac{1}{(2n+2)(2n+4)}\\ &> s_n+\frac14\dfrac{1}{(n+2)(n+2)}\\ &> s_n+\frac14(\dfrac1{n+1}-\dfrac1{n+2})\\ \end{array} $

so that $s_{n+1}-s_n \gt \frac14(\dfrac1{n+1}-\dfrac1{n+2}) $.

Summing, $\sum_{k=1}^{m-1}(s_{n+k+1}-s_{n+k}) \gt \sum_{k=1}^{m-1}\frac14(\dfrac1{n+1}-\dfrac1{n+2}) $ or $s_{n+m}-s_n \gt \frac14(\frac1{n+1}-\frac1{n+m+1}) $ so that $S =\lim_{m \to \infty}s_{n+m} \ge s_n+\lim_{m \to \infty}\frac14(\frac1{n+1}-\frac1{n+m+1}) = s_n+\frac1{4n+4} $.

Therefore $S \ge s_n +\dfrac1{4n+4}$.

Therefore, for any $n$, $s_n +\dfrac1{4n+4} \le S < s_n +\dfrac1{4n}$.