Let $x>0$ be a positive real number and the series $$\sum_{n=1}^\infty \left\lvert \frac{1}{(n+1)^x}-\frac{1}{n^x} \right\rvert$$ How can I proof that it is convergent?
Thanks
how to proof the Convergence of $\sum_{n=1}^\infty \left\lvert \frac{1}{(n+1)^x}-\frac{1}{n^x} \right\rvert$
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real-analysis
sequences-and-series
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3The terms are negative so the absolute value is $1/n^x-1/(n+1)^x$ and the partial sum telescopes so the sum is $1$ – 2017-01-29
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0@marwalix thanks so much – 2017-01-29
1 Answers
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Let $f(t)=\frac1{t^x}$. Then by the Lagrange MVT, one has $$ f(t+1)-f(t)=f'(\xi)=-\frac{x}{\xi^{x+1}},\xi\in(t,t+1) $$ and hence $$ \bigg|\frac{1}{(n+1)^x}-\frac1{n^x}\bigg|=|f(n+1)-f(n)|=f'(\xi)=\bigg|\frac{x}{\xi^{x+1}}\bigg|\le \frac{x}{n^{x+1}},\xi\in(n,n+1). $$ Since $x>0$, one has that $\sum_{n=1}^\infty\frac{x}{n^{x+1}}$ converges, which implies that $\sum_{n=1}^\infty\bigg|\frac{1}{(n+1)^x}-\frac1{n^x}\bigg|$ conerges.