Series $\sum\limits_{n=0}^{\infty}\frac{n+1}{2^n}$

Question

I want to check, whether $\sum\limits_{n=0}^{\infty}\frac{n+1}{2^n}$ converges or diverges.

I tried to use the Ratio test:

$|\frac{a_{n+1}}{a_n}|= \frac{n+2}{2^{n+1}} \frac{2^n}{n+1} = \frac{1}{2} \frac{n+1+1}{n+1} = \frac{1}{2} (1+ \frac{1}{n+1})$

$\lim\limits_{n \rightarrow \infty}{{(\frac{1}{2}(1+ \frac{1}{n+1})) = \frac{1}{2}}} \leq 1$

So $\sum\limits_{n=0}^{\infty}\frac{n+1}{2^n}$ converges.

Could somebody please check my solution?

Answer 1

It looks fine to me. Perhaps you'd want to write it a bit neater but everything is there and the 'correct' test has been used. (the $\le$ doesn't matter so much since your limit is less than $1$ strictly. However, you should really have $<$).

Answer 2

You can also break it down into a sum of two series $1/2^n $ and $n/2^n$ , each of which converges; first is a geometric series (with $r<1$), the second converges by , e.g., the zeta test; $n/2^n < 1/n^2$ for all $n>N$ ; $N$ a finite value.

Answer 3

The series converges, it's can be simply proved finding it's upper bound or as you've done it by the ratio test. But if you are interested in finding it's actual value here's a way to do it. From the sum of geometric series we have:

$$\sum_{n=0}^{\infty} r^n = \frac{1}{1-r}$$

Multiply both sides by $r$

$$\sum_{n=0}^{\infty} r^{n+1} = \frac{r}{1-r}$$

Differentiate with respect to $r$. Then we have:

$$\sum_{n=0}^{\infty} (n+1)r^n = \frac{1}{(r-1)^2}$$

Now substitute $r= \frac 12$. We have:

$$\sum_{n=0}^{\infty} \frac{n+1}{2^n} = \frac{1}{(-\frac 12)^2} = 4$$

Answer 4

$\sum_{n=0}^\infty (n+1) 2^{-n}$

$ = \sum_{n=0}^\infty \left( n 2^{-(n-1)} - (n+1) 2^{-n} + 2 \cdot 2^{-n} \right)$

$ = \left( 0 \cdot 2^{-(0-1)} - \lim_{n \to \infty} (n+1)2^{-n} \right) + 2 \sum_{n=0}^\infty { 2^{-n} }$

$ = 4$

The first step can be found be guesswork in simple cases, or be systematically doing "anti-difference by parts" in almost the same manner as "integration by parts", where the "by-parts" formula can be derived in exactly the same way from the "product rule" for the difference operator. Either of the forward or backwards difference operators work. This can easily handle things like $\sum_{n=0}^\infty n^3 3^{-n}$.