Question Regarding Arithmetico-Geometric Sequences

Question

Question: Is there a way to show that the well-known Arithmetico-Geometric Sequence$$\sum\limits_{n\geq0}\frac n{2^n}=2$$Using recurrence relations?

I'm well aware that there is an explicit formula for these kinds of series, but I'm wondering if solving these kinds of series using recurrence relations can be done. Recurrence equations work for geometric and arithmetic sequences, so why not for these hybrids?

For the sum in the question, I got the recurrence$$u_n=\frac 12u_{n-1}+\frac 1{2^n}\qquad u_0=0$$But I keep on getting zero when I take the limit as $n\to\infty$. Here's how I solved it. The homogenous solution for the equation is given as$$u_n^{(h)}=C_1\left(\frac 12\right)^n$$and the particular solution is give as$$u_n^{(p)}=C_2n\left(\frac 12\right)^n+C_3n$$Substituting that into the equation, and letting $n=0,1$, I get the two systems of equations$$C_2+C_3=1$$$$C_2+2C_3=1$$So $C_3=0$ and $C_2=1$. Of course, this doesn't make much sense because then $C_1$ must equal zero. And that would mean that as $n\to\infty$, our sum equates to zero.

It's also been a while since I've done some solving in recurrence relations.

Answer 1

If you really want to use a recurrence relation, then it helps to write out the whole sum like so$$\frac 01+\frac 12+\frac 24+\frac 38+\frac 4{16}+\cdots+\frac n{2^n}$$From there, since you want to get the sum, the recurrence relation can be written as$$u_n=u_{n-1}+\frac n{2^n}\tag1$$In fact, the recurrence for the first $n$ terms of a sequence of numbers will always be the $n-1$ term added to the $n$th term. From here, it's obvious that the homogeneous solution of (1) is given as$$u_n^{(h)}=C_1\cdot1^n=C_1\tag2$$And the particular solution is given by taking the general case of the remaining parts. Therefore$$u_n^{(p)}=C_2n\left(\frac 12\right)^n+C_3\left(\frac 12\right)^n$$Of which, the constants $C_2$ and $C_3$ can be directly substituted back into (1) to get$$u_n^{(p)}=-n\left(\frac 12\right)^n-2\left(\frac 12\right)^n\tag3$$Adding (2) and (3) together, and since $u_0=0$, we have the final recurrence as$$\large\boxed{u_n=2-\frac {n+2}{2^n}}$$When $n\to\infty$, the sum evaluates to $2$$$\large\boxed{\sum\limits_{n\geq0}\frac n{2^n}=\lim\limits_{n\to\infty}\left[2-\frac {n+2}{2^n}\right]=2}$$

Answer 2

Your recurrence relation is for the terms of the series, not the partial sums. So of course the limit is zero. You can derive the sum using the recurrence relation in the following way. We have for $n\geq 1$, $u_n=\frac12 u_{n-1} + \frac1{2^n}$. Therefore:

$$\sum_{n\geq1} u_n=\frac12\sum_{n\geq1} u_{n-1}+\sum_{n\geq1}\frac1{2^n}$$

Since $u_0=0$, the first two sums in that equation are equal. If $S$ is that sum, this is just a linear equation for $S$, which when solved yields $S=2$.

Answer 3

Consider the sequence of partial sums $$ S_{n}=\sum_{k=1}^{n}\frac{k}{2^{k}}. $$ Since $$ 2S_{n}=\sum_{k=1}^{n}\frac{k}{2^{k-1}}=\sum_{k=0}^{n-1}\frac{k+1}{2^{k}}=\sum_{k=0}^{n-1}\frac{k}{2^{k}}+\sum_{k=0}^{n-1}\frac{1}{2^{k}}=S_n-\frac{n}{2^{n}}+\dfrac{1-\frac{1}{2^{n}}}{1-\frac{1}{2}}=S_n-\frac{n}{2^{n}}+2-\frac{2}{2^{n}}, $$ we get $$ S_{n}=2S_{n}-S_{n}=2-\frac{n}{2^{n}}. $$ Hence $$ \sum_{n=1}^{\infty}\frac{n}{2^{n}}=\lim_{n\to\infty}S_{n}=\lim_{n\to\infty}\left(2-\frac{n}{2^{n}}\right)=2 $$