MHB Calculate Limit of Series: Step-by-Step Guide

[esuahcdss12]

hey
I am trying to calculate the limit of :

limn→∞(1/2+3/4+5/8+...+2n−1/2^n)
but I am not sure how to solve it, I thought to calculate 2S and than subtract S, but it did not worked well. I did noticed that the denominator is a geometric serie,but I don't know how to continue. could you help?

 

[tkhunny]

It should have worked well! Your thinking is good.

You are on a very successful path. Perhaps you just didn't quite interpret your results correctly.

Please demonstrate your efforts.

What does 2S - S look like?

 

[MarkFL]

As suggested, I would continue on your current path, but I wanted to show how I would approach the problem...

I would begin by expressing the partial sum $S_n$ in the following difference equation:

$$S_{n}-S_{n-1}=(2n-1)2^{-n}$$

We see the homogeneous solution is:

$$h_n=c_1$$

And the particular solution will have the form:

$$p_n=(An+B)2^{-n}$$

Substituting $p_n$ into our difference equation, there results:

$$(An+B)2^{-n}-(A(n-1)+B)2^{-(n-1)}=(2n-1)2^{-n}$$

Multiply through by $2^{-n}$:

$$(An+B)-2(A(n-1)+B)=2n-1$$

$$An+B-2(An-A+B)=2n-1$$

$$An+B-2An+2A-2B=2n-1$$

$$-An+2A-B=2n-1$$

Equating like coefficients, we obtain:

$$-A=2\implies A=-2$$

$$2A-B=-1\implies B=-3$$

And so our particular solution is:

$$p_n=-(2n+3)2^{-n}$$

And hence, the general solution is:

$$S_n=h_n+p_n=c_1-(2n+3)2^{-n}$$

Using the initial value, we find:

$$S_1=c_1-(2+3)2^{-1}=\frac{1}{2}\implies c_1=3$$

And so the solution satisfying all given conditions is:

$$S_n=3-(2n+3)2^{-n}$$

Thus, we find:

$$S_{\infty}=\lim_{n\to\infty}S_n=3$$

 

[esuahcdss12]

I manged to solve the question, i made an arithmetic error.
thanks