# Thread: testing series for conv/div.

1. ## testing series for conv/div.

how do i find if a series converges/diverges for this type of series:

inf
Σ [(-1)^k]*[(2^k)/ln (k)]
k=2

2. ## Solution

I'm wondering if you could use the ratio test for this:
|an+1/an|, So you would find the an+1 term:

[(-1)^(k+1)]*(2^(k+1))/ln(k+1)

Now, do the ratio:
[[(-1)^(k+1)]*(2^(k+1))/ln(k+1)] / [((-1)^k)*(2^k)/ln(k)]

Can be rewritten as:
[[(-1)^(k)*(-1)]*(2^(k)*2))*ln(k)] / [ln(k+1)((-1)^k)*(2^k)]

Now, you can cancel the (-1)^k term and the 2^k term from the numerator and the denominator to get:
(-1)*2*ln(k) / (ln(k+1))

Remember, you take the absolute value of the ratio so it reduces to:

2lnk/ln(k+1)

Now I don't know how to show that as k --> infinity, ln(k) and ln(k+1) are essentially equal. Then, you would be left with a ratio of 2. Since 2>1, the series would be said to diverge. I don't really know if this helps or not.

3. Originally Posted by VorpalX7
how do i find if a series converges/diverges for this type of series:

inf
Σ [(-1)^k]*[(2^k)/ln (k)]
k=2
If this is $\displaystyle \sum_{n=2}^{\infty}\frac{(-1)^n2^n}{\ln(n)}$ consider that $\displaystyle \lim_{n\to\infty}\frac{(-1)^n2^n}{\ln(n)}\ne{0}$ therefore the series diverges by the n-th term test