Could you help me show convergence/divergence of such series?

Question

Let a_n, n =1,2,3... be a sequence of positive real numbers and let k >0. Let s_n = a_1 + a_2....+ a_n be the sequence of partial sums of a_n. Show that the series

1) Sum (a_n)/(a_n + k) and 2) Sum (a_n)/(s_n) converges if, and only if, s_n converges.

Thank you for any help.

Answer 1

1) Suppose a_n converges. Since k and the terms a_n are positive, for every n, we have 0 < (a_n)/(a_n + k) < (a_n_)/k (1). Since Sum a_n converges, then so does Sum (a_n)/k. By comparison, (1) shows Sum (a_n)/(a_n + k) converges and, in addition, that Sum (a_n)/(a_n + k) <= Sum(a_n)/k.

For the converse, suppose Sum (a_n)/(a_n + k) converges. Then lim (a_n)/(a_n + k) = lim (1 - k/(a_n + k)) = 1 - lim k/(a_n + k)) = 0, which implies that k/(a_n + k) = 1. So, lim(a_n + k)/k = lim a_n/k + 1 = 1, so that lim a_n = 0. It follows that lim (a_n)/((a_n)/(a_n + k)) = lim a_n + k= k >0. By the limit comparison test, it follows the series Sum (a_n)/(a_n + k) and Sum a_n are both convergent or both divergent. Since the 1st converges, it follows Sum a_n also converges. This completes the proof.

2) Suppose s_n converges. Then, since the terms a_n are positive, lim s_n >0 and lim (a_n)/((a_n)/(s_n)) = lim s_n >0. So, by the limit comparison test, s_n and Sum (a_n)/(s_n) are both convergent or divergent. Since s_n converges, it follows Sum (a_n)/(s_n) also converges.

For the converse, suppose s_n diverges. Since the terms a_n are positive, it follows s_n --> oo. Let t_n be the sequence of partial sums of (a_n)/(s_n). For every n < m we have 0 < t_m - t_n = (a_(n +1))/(s_(n+1)) ....+ (a_m)/(s_m) < (a_(n +1))/(s_m) ....+ (a_m)/(s_m), because s_n is strictly increasing since the terms a_n are positive. So, for every n < m, we have

0 < t_m - t_n < (s_m - s_n)/(s_m) = 1 - (s_n)/(s_m)

If we keep n fixed and let m --> oo, then, since s_m --> oo, the right hand side goes to 1. So, for every n, we can find m > n such that t_m - t_n > 1/2. By the Cauchy criterion for series, it follows t_n diverges, that is, Sum (a_n)/(s_n) diverges.

The proof is now complete.

Answer 2

Some things to start you off...

1)
0 < (a_n)/(a_n+k) < (a_n)/k.

If Sum (a_n)/(a_n + k) converges, then a_n approaches 0. For sufficiently large n: a_n < k, and so (a_n)/(a_n + k) > (a_n)/(k + k).

2)
(a_n)/(a_1) <= (a_n)/(s_n) <= (a_n)/(lim s_n)

Answer 3

for (a_n)/(s_n) to converge s_n should be comparable with a_n
this is possible when the series is such that higher elements are
in a position of neglection like 1/2,1/4,1/16,1/32....... this means that sum is a converging one.