How do you find the sum of an infinite series?

[goohu]

Hi, I'm trying to solve the sum of following infinite series:

\(\displaystyle \sum_{k=1}^{\infty} \frac{{k}^{2}+4}{{2}^{k}} = \sum_{k=1}^{\infty} \frac{{k}^{2}}{{2}^{k}}
+ \sum_{k=1}^{\infty} \frac{4}{{2}^{k}}\)

Using partial sum we can rewrite the first series: \(\displaystyle \sum_{k=1}^{\infty} \frac{{k}^{2}}{{2}^{k}} = \sum_{k=1}^{n} \frac{{k}^{2}}{{2}^{k}} = \frac{n(n+1)(2n+1)}{6} * \left(\frac{1}{2}\right)^{\!{k}} \) , as n goes to \(\displaystyle \infty\)

The second series is a geometric one and can be rewritten as: \(\displaystyle \sum_{k=1}^{\infty} \frac{4}{{2}^{k}} = 4* \sum_{k=1}^{\infty} \left(\frac{1}{2}\right)^{\!{k}} = 1/2 \)

It's unclear how to get the sum from the term \(\displaystyle \frac{n(n+1)(2n+1)}{6} \)

 

[topsquark]

Hi, I'm trying to solve the sum of following infinite series:

\(\displaystyle \sum_{k=1}^{\infty} \frac{{k}^{2}+4}{{2}^{k}} = \sum_{k=1}^{\infty} \frac{{k}^{2}}{{2}^{k}}
+ \sum_{k=1}^{\infty} \frac{4}{{2}^{k}}\)

Using partial sum we can rewrite the first series: \(\displaystyle \sum_{k=1}^{\infty} \frac{{k}^{2}}{{2}^{k}} = \sum_{k=1}^{n} \frac{{k}^{2}}{{2}^{k}} = \frac{n(n+1)(2n+1)}{6} * \left(\frac{1}{2}\right)^{\!{k}} \) , as n goes to \(\displaystyle \infty\)

The second series is a geometric one and can be rewritten as: \(\displaystyle \sum_{k=1}^{\infty} \frac{4}{{2}^{k}} = 4* \sum_{k=1}^{\infty} \left(\frac{1}{2}\right)^{\!{k}} = 1/2 \)

It's unclear how to get the sum from the term \(\displaystyle \frac{n(n+1)(2n+1)}{6} \)

A quick note:
\(\displaystyle \sum_{k = 1}^{\infty} \dfrac{k^2}{2^k} \neq \sum_{k = 1}^{\infty} k^2 \cdot \sum_{k = 1}^{\infty} \dfrac{1}{2 ^k} \)

-Dan

 

[goohu]

Can I get hints? I've no idea how to continue from there

 

[HallsofIvy]

Hi, I'm trying to solve the sum of following infinite series:

\(\displaystyle \sum_{k=1}^{\infty} \frac{{k}^{2}+4}{{2}^{k}} = \sum_{k=1}^{\infty} \frac{{k}^{2}}{{2}^{k}}
+ \sum_{k=1}^{\infty} \frac{4}{{2}^{k}}\)

Using partial sum we can rewrite the first series: \(\displaystyle \sum_{k=1}^{\infty} \frac{{k}^{2}}{{2}^{k}} = \sum_{k=1}^{n} \frac{{k}^{2}}{{2}^{k}} = \frac{n(n+1)(2n+1)}{6} * \left(\frac{1}{2}\right)^{\!{k}} \) , as n goes to \(\displaystyle \infty\)

I presume that last term was supposed to be \(\displaystyle \left(\frac{1}{2}\right)^n\). Also, \(\displaystyle \sum_{k=1}^\infty

\frac{{k}^{2}}{{2}^{k}}​

\)

is NOT equal to \(\displaystyle \sum_{k=1}^n

\frac{{k}^{2}}{{2}^{k}}​

\)

. The first is the limit of the latter as n goes to infinity.
​

​

The second series is a geometric one and can be rewritten as: \(\displaystyle \sum_{k=1}^{\infty} \frac{4}{{2}^{k}} = 4* \sum_{k=1}^{\infty} \left(\frac{1}{2}\right)^{\!{k}} = 1/2 \)

It's unc+lear how to get the sum from the term \(\displaystyle \frac{n(n+1)(2n+1)}{6} \)

 

[skeeter]

\(\displaystyle \sum_{k=1}^\infty \frac{k^2+4}{2^k} = \sum_{k=1}^\infty \frac{k^2}{2^k} + \sum_{k=1}^\infty \frac{4}{2^k}\)

the last summation on the RHS is just the infinite geometric series ...

\(\displaystyle \sum_{k=1}^\infty \frac{4}{2^k} = 4\sum_{k=1}^\infty \left(\frac{1}{2}\right)^k = 4 \cdot \frac{\frac{1}{2}}{1-\frac{1}{2}} = 4 \cdot 1 = 4\)for the first summation on the RHS (saw this done a long time ago, so bear with me) ...

\(\displaystyle \sum_{k=1}^\infty \frac{k^2}{2^k} = \frac{1}{2} + \frac{4}{4} + \frac{9}{8} + \frac{16}{16} + \frac{25}{32} + \,...\)

\(\displaystyle 2 \sum_{k=1}^\infty \frac{k^2}{2^k} = 1 + \frac{4}{2} + \frac{9}{4} + \frac{16}{8} + \frac{25}{16} + \,...\)

\(\displaystyle 2 \sum_{k=1}^\infty \frac{k^2}{2^k} - \sum_{k=1}^\infty \frac{k^2}{2^k} = 1 + \frac{3}{2} + \frac{5}{4} + \frac{7}{8} + \frac{9}{16} + \frac{11}{32} + \, ...\)

\(\displaystyle \sum_{k=1}^\infty \frac{k^2}{2^k} - 1 = \frac{3}{2} + \frac{5}{4} + \frac{7}{8} + \frac{9}{16} + \frac{11}{32} + \, ...\)

\(\displaystyle 2\left(\sum_{k=1}^\infty \frac{k^2}{2^k} - 1\right) = 3 + \frac{5}{2} + \frac{7}{4} + \frac{9}{8} + \frac{11}{16} + \, ...\)

\(\displaystyle 2\left(\sum_{k=1}^\infty \frac{k^2}{2^k} - 1\right) - \left(2 \sum_{k=1}^\infty \frac{k^2}{2^k} - \sum_{k=1}^\infty \frac{k^2}{2^k}\right) = 2 + \frac{2}{2} + \frac{2}{4} + \frac{2}{8} + \frac{2}{16} + \, ...\)

\(\displaystyle \sum_{k=1}^\infty \frac{k^2}{2^k} - 2 = 2 + \frac{2}{2} + \frac{2}{4} + \frac{2}{8} + \frac{2}{16} + \, ... = 2\left(1 +\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \, ... \right) = 2(2) = 4\)

\(\displaystyle \sum_{k=1}^\infty \frac{k^2}{2^k} = 6\)

 

[topsquark]

Nicely done! And very innovative.

I tried this one out on W|A but it wouldn't give the steps. It did come up with a partial sum that looks like it fell from outer space. I have no idea how the partial sum was got.

This reminds me of a number of high level Math proofs in Physics. First you define the problem. You work it out a ways. Then you redefine the problem. Rinse and repeat until you get something you know how to solve. (The combination of Green's functions and Fourier transforms still seems like the blackest of magics to me.)

-Dan

 

[skeeter]

Nicely done! And very innovative.

I tried this one out on W|A but it wouldn't give the steps. It did come up with a partial sum that looks like it fell from outer space. I have no idea how the partial sum was got.

This reminds me of a number of high level Math proofs in Physics. First you define the problem. You work it out a ways. Then you redefine the problem. Rinse and repeat until you get something you know how to solve. (The combination of Green's functions and Fourier transforms still seems like the blackest of magics to me.)

-Dan

I didn't come up with this on my own ... saw it at an AP Calculus workshop one summer where the presenter showed how to complete some difficult looking summations w/o calculus. Took me a while (and about 5-6 pages in my composition book) to recall the solution.

 

[Opalg]

Another way to calculate the sum \(\displaystyle \sum_{k=1}^\infty \frac{k^2}{2^k}\) is to start with the geometric series $$\sum_{k=1}^\infty \frac{x^k}{2^k} = \frac{\frac x2}{1-\frac x2} = \frac{x}{2-x}.$$ Differentiate both sides to get $$\sum_{k=1}^\infty \frac{kx^{k-1}}{2^k} = \frac2{(2-x)^2}.$$ Multiply both sides by $x$: $$\sum_{k=1}^\infty \frac{kx^k}{2^k} = \frac{2x}{(2-x)^2}.$$ Now differentiate again. After a bit of simplification, that becomes $$\sum_{k=1}^\infty \frac{k^2x^{k-1}}{2^k} = \frac{4+2x}{(2-x)^3}.$$ Finally, put $x=1$ and you get \(\displaystyle \sum_{k=1}^\infty \frac{k^2}{2^k} = 6.\)

In general, differentiating an infinite series of functions term by term is not a valid procedure. But there is a theorem which says that a power series can be differentiated term by term within its radius of convergence. The above series all have radius of convergence $2$, so the differentiation is valid in the region of $x=1$.