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Let $\{ a_k \}_{k = 1}^{\infty}$ be a non-increasing sequence of non-negative reals such that $\displaystyle \sum_{k = 1}^{\infty} a_k = 1$. Define $\displaystyle T_k := \sum_{l = k+1}^{\infty} a_l$. I am trying to show that the ratio $\dfrac{T_k}{k a_k}$ is bounded from above by a constant, or in other words, it does not grow with $k$.

It seems to be true for polynomially and exponentially decaying sequences, but I can't seem to find a proof for the general case (or a counterexample to disprove the claim). Any hints or helps will be highly appreciated. Thanks!

sudeep5221
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2 Answers2

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Define $\displaystyle a_k = \frac{c}{k(\log (k+1))^2}$, where $$c := \left(\sum \frac{1}{k(\log (k+1))^2} \right)^{-1}.$$

Then $ka_k = c(\log(k+1))^{-2}$, while $$ T_k = \sum_{j \ge k+1} a_j \ge \int_{k+1}^{\infty} \frac{c}{x(\log (x+1))^2} \ \mathrm{d}x \ge \int_{k+1}^{\infty} \frac{c}{(x+1)(\log (x+1))^2} \ \mathrm{d}x = \frac{c}{\log(k+2)}. $$

Hence $$\frac{T_k}{ka_k} \ge c\frac{(\log(k+1))^2}{\log(k+2)} \to \infty$$ as $k \to \infty$.

fwd
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  • That's a neat solution! For my application, I can actually allow a growth upto a log factor. But based on your answer, one can easily extend it to any power of log which wouldn't be allowed in my original problem. Thanks! – sudeep5221 Sep 14 '23 at 18:50
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You can artificially make the ratio between $T_k$ and $ka_k$ arbitrarily large by suddenly letting $a_k$ dramatically shrink. And yes, it has to shrink super-exponentially fast. Here is one such construction.

For each positive integer $i$, let $n_i = 1+\sum_{j=0}^{i-2} 10^{2^j}$ (so $n_1 = 1$). Then for each $k$, $$a_k = \begin{cases} 2^{-i}*10^{-2^{i-1}} & \text{ if } k = n_i,\\ a_{n_i} &\text{ if } n_i < k < n_{i+1}. \end{cases}$$

Then it's easily verifiable that $a_k$ is non-increasing, $\sum_{k=1}^\infty a_k = 1$ and that for each $i$, $T_{n_i} + a_{n_i} = 2^{-i}$ so for $i > 1$,

\begin{align*} \frac{T_{n_i}}{n_ia_{n_i}} &= \frac{2^{-i}}{\left(1+\sum_{j=0}^{i-2} 10^{2^i}\right)2^{-i}10^{-2^{i-1}}} - \frac{1}{n_i}\\ &\geq \frac{2^{-i}}{2*10^{2^{i-2}}2^{-i}10^{-2^{i-1}}} - \frac{1}{n_i}\\ &= \frac{10^{2^{i-2}}}{2} - \frac{1}{n_i}\\ &\to \infty. \end{align*}

That proves that for the sequence $\{a_k\}$, $\frac{T_k}{ka_k}$ is unbounded.

forgottenarrow
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  • Thanks for the answer! I tried using a similar argument for the super-exponential case but must have missed something. Thanks for outlining the argument clearly! – sudeep5221 Sep 14 '23 at 18:48