If $A_n = \sum_{k = 1}^n \dfrac{k^6}{2^k}$, find $\lim_{n \to \infty} A_n$

Question

Set $$A_n = \sum_{k = 1}^n \dfrac{k^6}{2^k}.$$ Find $\displaystyle \lim_{n \to \infty} A_n$.

I tried solving this using a reduction method. That is, reducing the above series to an Arithmetico-Geometric series. We can see this by noting that $$\sum_{k = 1}^{\infty} \dfrac{k^m}{2^k} = \dfrac{2}{3} \left(1+\dfrac{2^m}{2} +\sum_{k=1}^{\infty} \dfrac{(k+1)^m-(k-1)^m}{2^k}\right)$$ which can be derived using infinite series. Using this method, we can eventually reduce the series to something of the form $\displaystyle C\sum_{k=1}^{\infty} \dfrac{k}{2^k}$ for some constant $C$ which is easily evaluated. Although this method would require a minimum of six iterations which would be quite lengthy. I am wondering if there is a more efficient way of solving this.

Many approaches exist! See http://math.stackexchange.com/q/439062/215011 and http://math.stackexchange.com/q/1681975/215011 for a few examples. — grand_chat, Apr 28 '16 at 18:46

score 0 · Answer 1 · answered Apr 28 '16 at 17:47

0

Note that $$\sum_{k=1}^\infty x^k = \frac{1}{1-x} - 1$$ converges uniformly on every compact subset of $(-1,1)$ so that $$\sum_{k=1}^\infty k x^{k-1} = \frac{1}{(1-x)^2}$$ by differentiation and $$\sum_{k=1}^\infty kx^k = \frac{x}{(1-x)^2}$$ after multiplication by $x$.

Repeat five times and plug in $x=\frac 12$.

answered Apr 28 '16 at 17:47

Umberto P.

52,165

I don't see how this helps. This is just the formula for an Arithmetico-Geometric series. – user19405892 Apr 28 '16 at 17:52
@user19405892 Could be worse! It gives $\frac{x^6+57x^5+302x^4+302x^3+57x^2+x}{(x-1)^7}$. Substituting $x=\frac{1}{2}$ gives 9366 as the answer. – almagest Apr 28 '16 at 17:59
@almagest Do you have to check that the series converges otherwise doing arithmetic operations might result in a contradiction? – user19405892 Apr 28 '16 at 18:57
All the series like $\frac{1}{1-x}$ converge for $|x|<1$. – almagest Apr 28 '16 at 18:58
@almagest Yes, but why is it necessary to say it converges as in the answer above? – user19405892 Apr 28 '16 at 18:59
Because we want the closed form to be the same as the infinite series when $x=\frac{1}{2}$. – almagest Apr 28 '16 at 19:00
@user19405892 But the algebra involved in getting the expression I gave above is horrendous. You may be correct that there is a better way of doing it. – almagest Apr 28 '16 at 19:01
@almagest How did he go from $$\sum_{k=1}^\infty x^k = \frac{1}{1-x} - 1$$ to $$\sum_{k=1}^\infty k x^{k-1} = \frac{1}{(1-x)^2}?$$ – user19405892 Apr 28 '16 at 19:15
@user19405892 By differentiating each side. – almagest Apr 28 '16 at 19:19

If $A_n = \sum_{k = 1}^n \dfrac{k^6}{2^k}$, find $\lim_{n \to \infty} A_n$

1 Answers1

Linked