Suppose that $a_n$ is some sequence, and let $f,g:\mathbb{N}\to\mathbb{N}$ such that $f(n) \leq g(n)$ for all $n$, and that $\lim\limits_{n\to\infty}\frac{g(n)}{f(n)} = 1$. Is it true that $$ \lim\limits_{n\to\infty}\sum\limits_{k=f(n)}^{g(n)}a_k = 0 $$ if $\lim\limits_{k\to\infty}a_k = 0$? The intuition for this question: eventually, the limits become the same, and therefore each element of the sequence $\sum\limits_{k=f(n)}^{g(n)}a_k$ is simply $a_{f(n)}$, and is getting smaller and smaller.

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    $\begingroup$ No!! It may be possible that $f(n)-g(n)$ becomes too big for $a_{n}$ to handle. Check with $a_{n}=1/\log n, f(n)=n,g(n)=n +[\sqrt{n}]$ $\endgroup$ – Paramanand Singh May 18 '16 at 5:11
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    $\begingroup$ The issue with intuitive logic is that if the value of $a_{k}$ is small then sum of a finite number of such terms will be small, but when number of terms $g(n)-f(n)$ itself tends to infinity then anything can happen. The fact that $g/f \to 1$ does not ensure that $g-f$ is bounded. $\endgroup$ – Paramanand Singh May 18 '16 at 5:18

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