Proving Absolute Convergence of a Recursive Sequence

[wany]

Homework Statement

Given the recursive sequence: a_k=(-1)^k(1+k\sin(\frac{1}{k}))^{-1}a_{k-1}, k>1 with a_1=1. Prove that \displaystyle\sum\limits_{k=1}^{\infty} a_k converges absolutely.

Homework Equations

ratio test, l'hospital's rule

The Attempt at a Solution

So I know how to do this problem using the approximation of sine, but we cannot use that.
So I use the ratio test to get \mathop {\lim }\limits_{k \to \infty } |\frac{1}{1+(k+1)\sin(\frac{1}{k+1})}|
Then I use l'hopital's rule several times and get it down to:
\frac{-cos(\frac{1}{k+1})}{k+1}
So taking the limit as k goes to infinity of this, I know that cos is bounded by -1,1 and 1/(k+1) goes to 0, so this would go to 0.

So since 0 < 1, we can say that this series converges absolutely by the ratio test.

 

[micromass]

I don't see where you applied l'hopital on this limit:

\lim_{k \rightarrow \infty } \left|\frac{1}{1+(k+1)\sin(\frac{1}{k+1})}\right|

You are only allowed to apply l'hopital when you're in a 0/0 or infinity/infinity situation...
To evaluate this limit, you'll have to use the following:

\lim_{x\rightarrow 0}{\frac{\sin(x)}{x}}=1.

 

[wany]

I see. Ok so I already proved that the limit as x goes to 0, of sin(x)/x goes to 1 so that is good. So I am trying to see how to get to the form of sin(x)/x from what we have, but I can't seem to do it. If I let x=\frac{1}{k+1}, then I know as k goes to infinity that this will go to 0. In addition therefore it will be sin(x). However, I am not sure how to get from <br /> \lim_{k \rightarrow \infty } \left|\frac{1}{1+(k+1)\sin(\frac{1}{k+1})}\right|<br />
to a usable form. Can we drop the one on the bottom to get:
<br /> \lim_{k \rightarrow \infty } \left|\frac{1}{(k+1)\sin(\frac{1}{k+1})}\right|<br /> which thus by replacement is <br /> \lim_{x \rightarrow 0 } \left|\frac{x}{\sin(x)}\right|<br />
But then this is the inverse,but since it is 1, it will be 1 again? But therefore this does not show that the series converges absolutely which is what we need to show.

Any help, and thank you for the help so far?

 

[micromass]

Well, first calculate \lim_{k\rightarrow +\infty}{(k+1)\sin{\frac{1}{k+1}}}.
If you've calculated that (and if it isn't equal to -1), then you can use

\lim_{k\rightarrow +\infty}{\left|\frac{1}{1+(k+1)\sin{\frac{1}{k+1}}}\right|}=\left|\frac{1}{1+\lim_{k\rightarrow +\infty}{(k+1)\sin{\frac{1}{k+1}}}}\right|

 

[wany]

Ok, so can I let x=\frac{1}{k+1} so therefore
\lim_{k\rightarrow +\infty} (k+1)\sin(\frac{1}{k+1})=\lim_{x\rightarrow 0} \frac{\sin(x)}{x}<br />
which is equal to 1. So now applying this to the other thing, we get 1/2. Correct? I know this is what I got when I used the approximation of sine, and since this is less than 1 it will converge absolutely.

 

[micromass]

Yes, the limit will indeed be 1/2...

 

[wany]

Alright cool thank you very much for your help.