# Any tips about the solution ??? using the Mathematical induction and thanks :)

## Jun 21, 2018

See below.

#### Explanation:

Before even starting with the induction proof, it looks like the expression has a useless term: we have

${\sum}_{k = 1}^{2 n} \left(\setminus \frac{1}{k \left(k + 1\right)} + {\left(- 1\right)}^{k}\right) = {\sum}_{k = 1}^{2 n} \setminus \frac{1}{k \left(k + 1\right)} + {\sum}_{k = 1}^{2 n} {\left(- 1\right)}^{k}$

Let's focus on the second sum. We are summing an even number of times (from $1$ to $2 n$) the powers of $- 1$, which sum to zero. For, example, we have

$n = 1 : {\left(- 1\right)}^{1} + {\left(- 1\right)}^{2} = - 1 + 1 = 0$

$n = 2 : {\left(- 1\right)}^{1} + {\left(- 1\right)}^{2} + {\left(- 1\right)}^{3} + {\left(- 1\right)}^{4} = - 1 + 1 - 1 + 1 = 0$

and so on. So, the expression is the same if we ignore the second sum. So, let's define the proposition $P \left(n\right)$ as

${\sum}_{k = 1}^{2 n} \setminus \frac{1}{k \left(k + 1\right)} \setminus \ge \frac{2}{3}$

and let's prove $P \left(n\right)$ for all $n \setminus \ge 1$

For $n = 1$, we have to prove that

${\sum}_{k = 1}^{2} \setminus \frac{1}{k \left(k + 1\right)} \setminus \ge \frac{2}{3}$

We can explicitly write the first two terms:

$\setminus \frac{1}{1 \left(1 + 1\right)} + \setminus \frac{1}{2 \left(2 + 1\right)} = \frac{1}{2} + \frac{1}{6} = \frac{4}{6} = \frac{2}{3} \setminus \ge \frac{2}{3}$

So, the base case $P \left(1\right)$ holds.

Now, if we assume that $P \left(n\right)$ is true for some $n \setminus \in \setminus m a t h \boldsymbol{N}$, we must show that $P \left(n + 1\right)$ holds as well.

We have that $P \left(n + 1\right)$ is the proposition

${\sum}_{k = 1}^{2 \left(n + 1\right)} \setminus \frac{1}{k \left(k + 1\right)} = {\sum}_{k = 1}^{2 n + 2} \setminus \frac{1}{k \left(k + 1\right)} \setminus \ge \frac{2}{3}$

${\sum}_{k = 1}^{2 n} \setminus \frac{1}{k \left(k + 1\right)} \setminus \ge \frac{2}{3}$
This means that $P \left(n + 1\right)$ only has two terms more than $P \left(n\right)$, and all terms are positive, since they are $\setminus \frac{1}{k \left(k + 1\right)}$.
So, the sum up to $2 n$ is already greater than $\frac{2}{3}$, and we are adding two more (positive) terms, so we have
${\sum}_{k = 1}^{2 \left(n + 1\right)} \setminus \frac{1}{k \left(k + 1\right)} \setminus \ge {\sum}_{k = 1}^{2 n} \setminus \frac{1}{k \left(k + 1\right)} \setminus \ge \frac{2}{3}$