# How do you find the sum of an infinite series?

• MHB
• goohu
In summary, using partial sum, the infinite series can be rewritten into two separate series: one that can be solved using the formula for the sum of squares, and another that is a geometric series. After solving for these two series, the sum of the original series can be obtained by adding the two results together. The solution involves redefining the problem multiple times until it can be solved using known methods, similar to certain high level mathematical proofs.
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 $$\displaystyle \frac{n(n+1)(2n+1)}{6}$$

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goohu said:
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

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

goohu said:
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 $$\displaystyle \frac{n(n+1)(2n+1)}{6}$$

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

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) ...

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

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

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

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

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

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

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

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

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

topsquark said:
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.

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 $x=1$.

## 1. How do you know if an infinite series has a sum?

An infinite series is said to have a sum if the sequence of partial sums converges to a finite value. This means that as you add more and more terms in the series, the total value will approach a specific number.

## 2. What is the formula for finding the sum of an infinite series?

The formula for finding the sum of an infinite series is S = a/(1-r), where S is the sum, a is the first term in the series, and r is the common ratio between consecutive terms.

## 3. How do you determine the convergence of an infinite series?

The convergence of an infinite series can be determined by using various tests, such as the ratio test, the root test, or the integral test. These tests help to determine if the series will converge or diverge.

## 4. Can an infinite series have a sum of infinity?

Yes, an infinite series can have a sum of infinity. This occurs when the sequence of partial sums does not converge to a finite value, but instead grows infinitely large as more terms are added.

## 5. Is it possible to find the exact sum of every infinite series?

No, it is not always possible to find the exact sum of an infinite series. Some series have no known formula for their sum, while others may have a sum that is irrational or cannot be expressed in a closed form.

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