Show that $\prod_{k=2}^{N} 1+\frac{1}{k^2} < 2$

Question

I tried doing this, but I got stuck.

I am supposed to show that:

$\prod_{k=2}^{N} \left(1+\frac{1}{k^2} \right)< 2$

I tested for N = 2

1.25 < 2

Then, assuming the statement is true, I have to prove that:

$\prod_{k=2}^{N+1} \left(1+\frac{1}{k^2} \right)< 2$

I tried expanding the assumption I made true, then I multiplied both sides by $1+\frac{1}{(N+1)^2}$

$(1+\frac{1}{2^2})\cdot(1+\frac{1}{3^2})\cdot\cdot\cdot(1+\frac{1}{N^2})$ < 2

$(1+\frac{1}{2^2})\cdot(1+\frac{1}{3^2})\cdot\cdot\cdot(1+\frac{1}{N^2})\cdot(1+\frac{1}{(N+1)^2}) < 2(1+\frac{1}{(N+1)^2})$

$\prod_{k=2}^{N+1} 1+\frac{1}{k^2} < 2(1+\frac{1}{(N+1)^2})$

But because $2(1+\frac{1}{(N+1)^2})$ is always larger than 2, I cannot show the statement I wanted to prove. Or in other words, knowing that $\prod_{k=2}^{N+1} 1+\frac{1}{k^2} < 2 < 2(1+\frac{1}{(N+1)^2})$ doesn't help because $2(1+\frac{1}{(N+1)^2})$ is not less than 2, making me not able to take any conclusions.

I tried putting it to wolframalpha to get an approximate answer as N approaches infinity, and it gave me an approximation of 1.838, so this question should be solvable.

Can someone help me?

Answer 1

Hint: Take the log of the product. Using the Taylor series for $\ln(1+x)$, we have $\ln(1+\frac{1}{k^2})\leq\frac{1}{k^2}$. Now use the infinite sum of $\frac{1}{k^2}$ as an upper bound (remember to exclude the term $k=1$).

Answer 2

Hint: For all $t\ge 0$, $1+t\le e^t$. Put $t=\frac{1}{k^2}$. Note that $\sum_{k=2}^{n}{\frac1{k^2}}\le \frac{\pi^2}{6}-1$

Answer 3

Hint: Prove by induction by strengthening the inequality.

Like you realized, we can't induct directly since $ 2 ( 1 + \frac{1}{(k+1)^2 }) > 2$.

However, the trick is to find a $ g(N) $ such that

• $ g(N) ( 1 + \frac{1}{(N+1)^2 }) < g(N+1)$,
• $ 1 + \frac{1}{(1+1)^2} < g(2) $
• $ g(N) < 2$.

With such a $g(N)$, the induction works:

$$ \prod_{k=2}^{N+1} ( 1 + \frac{1}{ k^2 }) < g(N)( 1 + \frac{1}{(N+1)^2 }) < g(N+1) < 2 .$$

There are several candidates for such a function. If you can't find one, you can reveal it below.

$g(k) = 2 - \frac{2}{ k+1} $.