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I was wondering whether this series is divergent:

$$\sum_{n=2}^{\infty}\frac{1}{n\log^{1+\sin(n)}(n)}$$

Using Cauchy's Condensation, we can say that:

$$\sum_{n=2}^{\infty}\frac{1}{n\log^{1+\sin(n)}(n)} \quad \text{diverges} \iff \sum_{n=1}^{\infty}\frac{1}{n^{1+\sin(n)}} \quad \text{diverges}$$

Can you help me going on? Thanks in advance.

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Note that $$\begin{align} \sum_{n=2}^{\infty}\frac{1}{n\log^{1+\sin(n)}(n)}&\geq \sum_{k=0}^\infty\sum_{(2k+1)\pi < n < (2k+2)\pi}\frac{1}{n\log^{1+\sin(n)}(n)} \\ &\geq \sum_{k=0}^\infty\sum_{(2k+1)\pi < n < (2k+2)\pi}\frac{1}{n\log(n)}\\ &\geq \sum_{k=0}^\infty \sum_{(2k+1)\pi < n < (2k+2)\pi} \frac{1}{(2k+2)\pi\log((2k+2)\pi)} \end{align} $$

Since $(2k+2)\pi - (2k+1)\pi = \pi >1$, for each $k$ there is at least one integer $n$ such that $(2k+1)\pi < n < (2k+2)\pi$, hence $$\begin{align} \sum_{n=2}^{\infty}\frac{1}{n\log^{1+\sin(n)}(n)}&\geq \sum_{k=0}^\infty \frac{1}{(2k+2)\pi\log((2k+2)\pi)} \end{align} $$ Since $\displaystyle \frac{1}{(2k+2)\pi\log((2k+2)\pi)}\sim \frac{1}{2\pi k\log(k)}$ and $\displaystyle \sum_{k}\frac{1}{k\log(k)}$ diverges, the last inequality implies $$\sum_{n=2}^{\infty}\frac{1}{n\log^{1+\sin(n)}(n)} = \infty$$

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