Analysis - How to show a Series is bounded above: lim[n->infty]1/(2n)!

[ItsLozPot]

I've got a maths question i have been working on for ages and i just can't figure it out. I have to use the fact, 2^m-1<m! for all m in the natural numbers (no proof required) to show that the following sequence of partial sums is bounded above by 2/3.
The sum as n goes from 1 to infinity of 1/(2n)! ?

I've tried induction, ratio tests, limit tests, comparison tests and pretty much everything else i can think of and i'm not getting anywhere. Someone please help me, Thanks in advance xoxo

 

[ItsLozPot]

Proving a series is bounded above.

Consider the infinite series:

The Sum as n goes from 1 to infinity of 1/(2n)!

I've proved that the sequence of partial sums of this series is increasing and now i need to show it's bounded above by 2/3. It tells me to use the fact that 2^m-1<m! for all m in the natural numbers.

I've tried induction and various different tests (ratio, comparison, limit) and induction and i'm not getting anywhere, please help me!!

Thanks in Advance

 

[ksdhart2]

Okay, so let's start from the given statements. Best I can tell, you're asked to find the value of this infinite series:

\(\displaystyle \displaystyle \sum _{n=1}^{\infty }\:\frac{1}{\left(2n\right)!}\)

You're then told that, for any natural number m:

\(\displaystyle 2^{m-1}<m!\)

Based on that, we can say that:

\(\displaystyle \dfrac{1}{2^{2n-1}}>\dfrac{1}{(2n)!}\) and thus \(\displaystyle \displaystyle \sum _{n=1}^{\infty }\:\frac{1}{2^{2n-1}} > \sum _{n=1}^{\infty }\:\frac{1}{\left(2n\right)!}\)

Now it's just a matter of determining what value the bigger series converges to and that will be an upper bound. Based on the given answer, we know it's 2/3, but how can we prove that? Well, my method would be to manipulate it so that it looks like a geometric series, which has a known convergent value (for |r| < 1 anyway).

\(\displaystyle \dfrac{1}{2^{2n-1}}=\dfrac{1}{2^{2n}}\cdot \dfrac{1}{2^{-1}}=2\cdot \dfrac{1}{2^{2n}}\)

Try continuing from here and see what you get.

 

[lookagain]

It tells me to use the fact that 2^m-1<m! for all m in the natural numbers.

That's not true.

\(\displaystyle 2^{m - 1} < m! \ \ \) for m > 2 for m an integer.