# Series question

• May 1st 2009, 08:16 PM
Pinkk
Series question
Evaluate

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

So I separated the two sums and also change the dummy variable using $n=k-2$ so I can eventually have two geometric series:

$2\sum_{n=0}^{\infty}\,\,(\frac{1}{3})^{n+2} + \sum_{n=0}^{\infty}\,\,(\frac{1}{2})^{n+2}$

Pulling out the constants within the sums, I can use $\frac{a}{1-r}$ for both sums:

$\frac{\frac{2}{9}}{1-\frac{1}{3}} + \frac{\frac{1}{4}}{1-\frac{1}{2}}$

Simplifying, I get:

$\frac{2}{9}\cdot\frac{3}{2}+\frac{1}{4}\cdot\frac{ 2}{1}=\frac{1}{3}+\frac{1}{2}=\frac{5}{6}$

However, my book states that the answer is $\frac{6}{5}$. Did I do something wrong, or is the book's answer key wrong? Any help would be appreciated, thanks.
• May 1st 2009, 09:11 PM
Gamma
I got what you got when i just did it myself, but I did it slightly differently. Then looking at yours, yeah i see no flaws. That typo is pretty close, it seems pretty likely that they just messed up.

You may check online though, a lot of times publishers will put the errata online, it may well mention the mistake there. Just a thought
• May 1st 2009, 09:38 PM
redsoxfan325
Yes, Maple confirms that the answer is indeed $\frac{5}{6}$.
• May 1st 2009, 09:51 PM
Pinkk
Thanks guys. I have one more question before I pass out from tiredness:

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

How would one go about solving this. I can "see" that the series will diverge since $|a_{N+1}|>|a_{N}|$ when $N \ge 0$, essentially as you go farther out the distance $a_{k}$ is from the x-axis increases. How can show that this series diverges without giving such a wordy argument; essentially, how can I evaluate it exactly either by finding an expression for $S_{n}$ or some other method.
• May 1st 2009, 09:56 PM
Gamma
It is a necessary condition for a series to converge that the limit as n goes to infinity of the terms go to 0. This is clearly not the case, therefore the series will not convege.
• May 1st 2009, 10:03 PM
Pinkk
Ah, I think I get it, so eventually you'd have it where:

$\lim_{k\to\infty}(-1)^{k+1}\cdot\lim_{k\to\infty}\frac{3k}{k+1}$, and since $(-1)^{k+1}$ always oscillates between -1 and 1, $\lim_{k\to\infty}a_{k}$ does not exist and therefore the series diverges.
• May 1st 2009, 10:06 PM
Gamma
bingo.

Another classic one is like $\sum_{n=0}^{\infty}sin(n)$.

Everyone pulls out all the stops to try to use one of the tests, but forgets to first check the limit of the terms actually is 0. That should always be your first test because it is so easy to do and could really save you a ton of time in the end.
• May 1st 2009, 10:11 PM
simplependulum
THe answer is 3ln2 - 3k , k is 1 or 0 so the series diverges