Determien the sums of the following series'.

1:$\sum_{k=0}^\infty \frac{2+(-1)^k}{3^k}$

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

3:$\sum_{k=0}^\infty \frac{1}{4k^2-1}$

4:$\sum_{k=0}^\infty \sum_{l=0}^k{k \choose l}\frac{1}{2^{k+l}}$

I have always been bad with infinite series', so I decided to practice them strongly for the next couple of days. I found an exercise in a textbook and kinda troubling with some of them.

Here's what I was thinking of doing.

1: I thought about dividing this into two sums, $\sum_{k=0}^\infty \frac{2}{3^k}+\sum_{k=0}^\infty \frac{(-1)^k}{3^k}$, which should be okay since they converge, right? Anyway, then I can see the geometric series there, giving me $3+\frac{3}{4}=3,75$. Is that correct?

2: I think that's what's called a telescope series? So I tried writing down some of the first numbers to see where it starts canceling each other out:

$\frac{1}{1}-\frac{1}{3}+\frac{1}{2}-\frac{1 }{ 4}+\frac{ 1}{ 3}-\frac{ 1}{ 5}+\frac{1 }{ 4}-\frac{1 }{6 } + ...$. Since $\frac{1}{n}$ converges to zero for $n\to \infty$, the sum of this infinite series is $\frac{3}{2}$?

About 3 and 4. I'm pretty much lost there. Can't seem to find an approach to find the sum.

Any tips?

  • 1
    $\begingroup$ On number three, try using partial fractions. On number four, try plugging in some values for $k$ to see what the inside sum is. You should notice a pattern $\endgroup$ – Brent Jun 16 '15 at 16:33


$$\begin{align}&(3)\;\;4k^2-1=(2k-1)(2k+1)\implies\frac1{4k^2-1}=\frac12\left(\frac1{2k-1}-\frac1{2k+1}\right)\\{}\\ &(4)\;\;\sum_{l=0}^k\binom kl\frac1{2^{k+l}}=\frac1{2^k}\sum_{l=0}^k\binom kl\frac1{2^l}=\frac1{2^k}\left(1+\frac12\right)^k\end{align}$$

  • $\begingroup$ I don't quite see how you transformed the equations in (4) yet. $\endgroup$ – Rab Jun 16 '15 at 16:52
  • $\begingroup$ @Rab : $$\sum_{k=0}^\infty \sum_{l=0}^k\binom kl\frac{1}{2^{k+l}}=\sum_{k=0}^\infty \frac1{2^k}\left(1+\frac12\right)^k=\sum_{k=0}^\infty\left(\frac34\right)^k=... $$ $\endgroup$ – Timbuc Jun 16 '15 at 17:49

$1-$ $$\sum_{k=0}^{\infty }\frac{2+(-1)^k}{3^k}=\sum_{k=0}^{\infty }\frac{2}{3^k}+\frac{(-1)^k}{3^n}=2\frac{1}{1-1/3}+\frac{1}{1+1/3}$$ $3-$ $$\sum_{k=0}^{\infty }\frac{1}{4k^2-1}=\sum_{k=0}^{\infty }0.5[\frac{1}{2k-1}-\frac{1}{2k+1}]=0.5[1/1-1/3+1/3-1/5+1/5+....]=\frac{1}{2}$$


$(1)$ Looks good to me.

$(2)$ You are correct that this is a telescoping series and your answer is correct (but the index should start at $k=1$, not $k=0$) You could also rewrite the series as $$\begin{align}\sum_{k=1}^\infty \frac{1}{k}-\frac{1}{k+2} = \sum_{k=1}^\infty \frac{1}{k}-\sum_{k=1}^\infty \frac{1}{k+2} \\ = \left(1+\frac{1}{2}+ \sum_{k=3}^\infty \frac{1}{k}\right)-\sum_{k=3}^\infty \frac{1}{k} \\ = 1+\frac{1}{2}\end{align}$$ to more rigorously see that the series converges to $\frac{3}{2}$.

$(3)$ Write the fraction $\frac{1}{4k^2-1}$ as $\frac{1}{(2k-1)(2k+1)}$, then use partial fraction decomposition to find constants $A,B$ where $\frac{1}{(2k-1)(2k+1)} = \frac{A}{2k-1}+ \frac{B}{2k+1}$ You may again need to invoke an argument about telescoping series.

For $(4)$, is that really the binomial coefficient $\binom{k}{l} = \frac{k!}{l!(k-l)!}$? I did not encounter that in my calc class unit on infinite series. However, start writing out a few terms of the series to get a handle on what the general behavior is. That is, fix $k=0$ and find a value. Then set $k=1$ to get a few more values, etc.


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