Question

(PICTURE OF QUESTION IS ATTACHED FOR EASE OF VISUALISATION)
Determine if the series converges or diverges. If it converges, find its sum.

(a) summation of (-1)^k . [2^(k-1) + 4^(k/2)]/3^(k-1) from k=1 to +infinity

(b) summation of [1/(ln k) - 1/(ln (k+1))] from k=2 to +infinity

Answer 1

Do you know the alternating series test for a)

Answer 2

You can re-write a) in the form:
\[\sum_{k=1}^{\infty}(-1)^kb_k\] and then you must show that \(b_k \ge b_{k+1}\) and that \[\lim_{k\rightarrow \infty}b_k=0\]

Then you'll be able to show that the series converges.

You can then split your sum into two components and show that you have two geometric series with \(|r|<1\) so you can find the sum

Answer 3

On second thought... if you just show that you have a sum of two geometric series (again for a) ) then you wouldn't need to use the alternating series test... sorry that is just extra work for nothing because you know it would converge since \(|r|<1\)

Answer 4

For b), you'll notice that when writing out the sum (with actual numbers plugged in), that will have a telescoping series. Then it suffices to show that all the intermediate terms will cancel out. Then, you just need to find the limit of this partial sum as \(k\rightarrow\infty\)

Answer 5

If you need more explanations, just ask

Answer 6

For (a), I think a simple comparison to a known, convergent/divergent geoemetric series would suffice.

Answer 7

Yeah you will just need the geometric series.

Answer 8

For b) Since you get a telescoping series and a finite limit of the partial sum, you'll get the sum of the series (and hence is convergent)