2
$\begingroup$

Given the series ${a_n}$ : $a_n= \sum _{k=1}^{n}{\frac { \left( -1 \right) ^{k}}{{3}^{k}+\sqrt {k}}}$ prove that ${a_n}$ converges.

So far this is what I've done: I've split this into 2 sub-series: $k=2n$ and $k=2n+1$

for $(k=2n)$ I get: $\sum _{k=1}^{n} \left( {3}^{k}+\sqrt {k} \right) ^{-1} = 0$ (as k goes to infinity the whole series converges to 0)

for $(k=2n+1)$ I get:

$\sum _{k=1}^{n}- \left( \left( {3}^{k} \right) ^{-1}+\sqrt {-k} \right) ^{-1}$ (which also converges to 0 as k approaches to infinity)

Since both subs-series of $a_n$ converge to the same value, $0$, $a_n$ converges, in particular to $0$.

I'm not sure if this is a tight enough proof since maybe by chance it works out here specifically. I also checked separate limits as k approaches $-\infty$. They both converge to $0$ for each sub-series. What am I missing? Thanks.

  • 1
    Don't write $k=2n$ when the letter $n$ is already in use. Additionally, don't write $\sum_{k=1}^n$ if you mean to only sum over odds or evens in particular. Lastly: (a) how can a sum of positive numbers equal $0$, and (b) how are you getting *negative* $k$ in your sum over odd $k$?2012-02-02

2 Answers 2

2

$ |a_n| \le \sum_{k=1}^n \frac{ 1}{3^k + \sqrt k} \le \sum_{k=1}^n \frac{1}{3^k - \frac 12 3^k} < 2 \sum_{k=1}^n \left( \frac 13 \right)^k $ which converges because this is a geometric series. Since $a_n$ converges absolutely it converges conditionally.

You could also use Leibniz's criterion, which states that if a series has leading term $(-1)^k b_k$ with $b_k \to 0$, then $\sum (-1)^k b_k$ converges. The fact that $b_k = \frac 1{3^k + \sqrt k} \to 0$ is clear enough.

EDIT : As the comments pointed out, I didn't remember at the time all the hypothesis of Leibniz's Criterion, which are that on top of what I said, the $b_k$'s must be, up to some $N$, non-negative and decreasing.

Hope that helps,

  • 0
    Okay guys, sorry if I didn't remember all the hypothesis. It's good that you pointed them out, but my point was mostly that you could use the criterion, not that I wanted to describe it completely.2012-02-02
1

You can use Leibniz's Rule for alternating series.

If $\{ a_n\}$ is a monotonic decrasing sucesion of numbers, the alternate series converges, and if $S$ is the sum, then, for every $n > 0$: $ 0 < (-1)^n (S-s_n) < a_{n+1}$ and $0 < S < a_1$

Thus your series converges, and you can calculate it's sum approximately with the first inequality.