SOLUTION OF ELEMENTARY PROBLEM

E 2558. Proposed by A. Torchinsky, Cornell University

Suppose that $\sum a_n$ is a divergent series of positive terms, and let $\mbox s_n=a_1+\cdots +a_n$ for $n=1,2,\dots$. For which values of p does the series $\sum a_n/{s_n^p}$ converge?

Solution by Elmer K. Hayashi. We prove a more general theorem from which we deduce that $\sum a_n/{s_n^p}$ converges if and only if p>1.

Theorem. Let f(x), for x>0, be any nonnegative, continuous, monotonically decreasing, real-valued function. If $\sum a_n$ is a divergent series of positive terms and if $s_n=a_1 +\cdots +a_n$ for $n=1,2,\dots$, then

and

Proof: Intuitively we reason that if u=s_n then du is analogous to s_n - s_n-1=a_n. Hence $\sum a_n f (s_n)$ probably behaves somewhat like $\int f(u) du$. Furthermore, if F(x) is any antiderivative of the continuous function f(x), then $\int_a^bf(u)du = F(b)-F(a)$. Thus a natural series with which to compare $\sum a_n f (s_n)$ is the telescoping series

$$\sum_{n=2}^\infty\{ F(s_n) - F (s_{n-1}) \}$$ (1)

since

$$\sum_{k=2}^n \{ F (s_k) - F(s_{k-1})\} = F(s_n) - F (s_1) = \int_{s_1}^{s_n} f(x) \;dx.$$ (2)

From equation (2), it is apparent that the series (1) converges if and only if the integral, $\int_{s_1}^\infty f(x)\;dx$, is convergent. Now, by the mean value theorem,

for some c_k between s_k-1 and s_k. Since f is monotonically decreasing, we have for $k=2,3,\dots$,

and

Using the Comparison test, we arrive at the conclusion of the theorem.

If we take $f(x) = x^{-p}, p \geq 0$, we conclude that $\sum a_n/{s_n^p}$ converges for p > 1 and $\sum a_n/{s_{n-1}^p}$ diverges for $0 \leq p \leq 1$. In general, it is not true that if $\sum a_n f(s_{n-1})$ is diverent, then $\sum a_n f (s_n)$ is also divergent. For example, if $f(x) = \frac{1}{x\;log\;x}$, a₁ = s₁ = 1 + e and a_n = s_n-1 exp (2^n-1) for n > 1, then $s_n = \frac{(exp(2^n)-1)}{(e-1)}$. The series $\sum a_n f(s_{n-1})$ is divergent, by our theorem, since $\int_2^\infty1(x\;log\;x)\;dx$ is divergent. However $\sum a_n f (s_n)$ behaves like $\sum 2^{-n}$ and therefore is convergent. However, in the special case $f(x) = x^{-p},\;p \leq 1$, it is true that $\sum a_nf(s_n) = \sum a_n/{s_n^p}$ does diverge.

It suffices to prove $\sum a_n/s_n$ diverges since $\frac{a_n}{s_n^p} \geq \frac{a_n}{s_n}$ for all sufficiently large n if $s_n \rightarrow \infty$ as $n \rightarrow\infty$ and $p\leq1$. By the comparison test, the divergence of $\sum \frac{a_n}{s_n}$ would then imply the divergence of $\sum a_n/{s_n^p}$ for all $p\leq1$. There are two cases to consider. Either $a_n \leq s_{n-1}$ for all sufficiently large n or otherwise there exist infinitely many n such that a_n > s_n-1. In the former case, we have for all sufficiently large n

But we know $\sum a_n/{s_{n-1}}$ is divergent, and hence, by the comparison test, $\sum a_n/s_n$ is also divergent. In the second case, we have for infinitely many n

Thus $\lim_{n \rightarrow \infty}(\frac{a_n}{s_n}) \neq 0$ and $\sum a_n/s_n$ is divergent in this case, too. This completes our proof.

 Elmer K. Hayashi

		Department of Mathematics

		Wake Forest University

		Winston-Salem, N. C. 27109