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If $\lim_{N\to \infty} \sum_{n=N}^{2N} c_n = 0$ do we have that $\sum_{n=1}^{\infty} c_n$ converges? At first this did not seem true($\sum_{n=N}^{2N} (-1)^n$ is $0$ when N is odd), but I've failed to find a proper counter-example. I've tried thinking in terms of partial sums to prove that is it true but have had no luck. Can someone help?

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Consider $c_n = \dfrac{1}{(n+1)\log_2(n+1)}$. Then, $0 < c_n \le \dfrac{1}{(N+1)\log_2(N+1)}$ for all $N \le n \le 2N$.

Thus, $0 < \displaystyle\sum_{n = N}^{2N}c_n \le \displaystyle\sum_{n = N}^{2N}\dfrac{1}{(N+1)\log_2(N+1)} = \dfrac{1}{\log_2(N+1)} \to 0$ as $N \to \infty$.

But, $\displaystyle\sum_{n = 1}^{\infty}c_n = \sum_{n = 1}^{\infty}\dfrac{1}{(n+1)\log_2(n+1)}$ diverges by the integral test.

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Let

\begin{align*} c_1 = c_2 &= \frac{1}{2}\\ & \\ c_3 = c_4 = c_5 = c_6 &= \frac{1}{4}\times\frac{1}{2} = \frac{1}{8}\\ & \\ c_7 = c_8 = c_9 = c_{10} = c_{11} = c_{12} = c_{13} = c_{14} &= \frac{1}{8}\times\frac{1}{3} = \frac{1}{24} \end{align*}

and so on. More precisely, let $c_n = \dfrac{1}{2^kk}$ where $k = \lfloor\log_2(n+1)\rfloor$.

Then if $S_m = \displaystyle\sum_{n=1}^ma_n$, we have

$$S_{2^{p+1}-2} = \sum_{n=1}^{2^{p+1}-2} c_n = \sum_{k=1}^{p}\sum_{n=2^k-1}^{2^{k+1}-2}c_n = \sum_{k=1}^{p}\sum_{n=2^k-1}^{2^{k+1}-2}\frac{1}{2^kk} = \sum_{k=1}^{p}2^k\frac{1}{2^kk} = \sum_{k=1}^p\frac{1}{k}.$$

As the harmonic series diverges, $S_{2^{p+1}-2}$ diverges. Therefore $\displaystyle\sum_{n=1}^{\infty}c_n$ diverges.

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