# Find the solutions to this inquality by induction

emyt

## Homework Statement

sum of i from i = 1 to n 1/i^2 <= 2

## The Attempt at a Solution

I have a solution, that for any n the sum from i to n of 1/i^2 < sum 1/i(i-1) = 1/(i-1) - 1/i .. et c but this is not inductive.. can I get any hints? thank you

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## Homework Statement

$$\sum_{i=1}^{n}\ \frac{1}{i^2}\ \le\ 2$$

## The Attempt at a Solution

I have a solution, that for any n the sum from i to n of 1/i^2 < sum 1/i(i-1) = 1/(i-1) - 1/i ... etc but this is not inductive.. can I get any hints? thank you
Here is your original expression. Click on it to copy & edit your Original Post, if you like.

$$\sum_{i=1}^{n}\ \frac{1}{i^2}\le 2$$

Certainly, 1 < 2, so it's true for the base case.

Assume that $$\sum_{i=1}^{n}\ \frac{1}{i^2}\le 2$$ is true for n≥1. Show that $$\sum_{i=1}^{n+1}\ \frac{1}{i^2}\le 2$$ is true.

You may also get some idea from looking at the sum for n=1, n=2, n=3, n=4, …

And/Or look at $$2\ -\ \sum_{i=1}^{n}\ \frac{1}{i^2}$$ for several values of n.

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...

## The Attempt at a Solution

I have a solution, that for any n the sum from i to n of 1/i^2 < sum 1/i(i-1) = 1/(i-1) - 1/i .. etc but this is not inductive.. can I get any hints? thank you
I can't figure out what you you're saying here.

Certainly, $$\frac{1}{i^2}<\frac{1}{i(i-1)}\ \text{ for }\ i\ge2\ .$$

$$\sum_{i=1}^{n}\ \frac{1}{i^2}=1+\sum_{i=2}^{n}\ \frac{1}{i^2}\ <\ 1+\sum_{i=2}^{n}\frac{1}{i(i-1)}=1+\sum_{j=1}^{n-1}\ \frac{1}{(j+1)j}$$

As it turns out, $$\sum_{j=1}^{\infty}\ \frac{1}{(j+1)j}=1\ ,$$ so this should be do-able.

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Wolfram Alpha gives the following result:
$$\sum_{j=1}^{n-1}\ \frac{1}{(j+1)j}=1-\frac{1}{n}\$$

Proving that this is true is pretty straight forward using induction.

Note:
$$\sum_{j=1}^{(n+1)-1}\ \frac{1}{(j+1)j}=\frac{1}{(n+1)n}+\sum_{j=1}^{n-1}\ \frac{1}{(j+1)j}\ ,\text{ and }\frac{1}{(n+1)n}=\frac{n+1-n}{(n+1)n}=\frac{n+1}{(n+1)n}-\frac{n}{(n+1)n}$$​