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I don't know wheteher I determine convergence properly, because it seems for me thath I have a problem in the end.

$$\sum_{n=1}^{\infty}\frac{(n+1)^{\beta}-n^{\beta}}{n^{\alpha}}$$

I want to assume the asymptotic similarity with $\sum\frac{1}{n^{\delta}}$, which is convergent for $\delta>1$.

$$\sum_{n=1}^{\infty}(\frac{(n+1)^{\beta}}{n^{\alpha}}-\frac{n^{\beta}}{n^{\alpha}})\\\frac{\beta}{\alpha}<1\Rightarrow \beta<\alpha$$

Additionally: $$(n+1)^{\beta}-n^{\beta}<n^{\alpha}$$ $$ f(n) =(n+1)^{\beta}-n^{\beta}-n^{\alpha} $$ Setting to $0$ and dividing by $n^{\beta}$:

$$ \frac{n+1}{n}^{\beta}-1-n^{\alpha-\beta}=0 \\ (\frac{n+1}{n})^{\beta}-n^{\alpha-\beta}=1 / \ln(\cdot) \\ \beta\ln(1+\frac{1}{n})-(\alpha-\beta)\ln(n) = 0 \\ \beta=\alpha\frac{\ln(n)}{\ln(n+1)}$$

I don't know, whether the answer is $\alpha>1$ and $\beta=\alpha\frac{\ln(n)}{\ln(n+1)}$ or whether I'm wrong

EDIT I used the derivative for the f(x) and got: $$\beta(n+1)^{\beta-1}-\beta n^{\beta-1}-\alpha n^{\alpha-1}=0 \\1+\frac{1}{n}-1-\frac{\alpha}{\beta}n^{\alpha-\beta-1+1}=0\\ \frac{1}{n}=\frac{\alpha}{\beta}n^{\alpha-\beta}\\ \alpha-\beta<0$$

Funny
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2 Answers2

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$$(n+1)^{\beta}=n^{\beta}\left(1+\frac1n\right)^\beta=n^{\beta } \left(1+\frac{\beta }{n}+O\left(\frac{1}{n^2}\right)\right)$$ $$(n+1)^{\beta}-n^{\beta}=n^{\beta } \left(\frac{\beta }{n}+O\left(\frac{1}{n^2}\right)\right)$$ $$\frac{(n+1)^{\beta}-n^{\beta}}{n^{\alpha}}\sim \beta n^{\beta-\alpha-1}$$

Claude Leibovici
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We can use that

$$(n+1)^{\beta}=n^{\beta}\left(1+\frac1n\right)^\beta=n^\beta+\beta \frac1{n^{1-\beta}}+\frac{\beta (\beta-1)}2\frac1{n^{2-\beta}}+O\left(\frac1{n^{3-\beta}}\right)$$

and therefore

$$\frac{(n+1)^{\beta}-n^{\beta}}{n^{\alpha}}=\beta \frac1{n^{1+\alpha-\beta}}+\frac{\beta (\beta-1)}2\frac1{n^{2+\alpha-\beta}}+O\left(\frac1{n^{3+\alpha-\beta}}\right)$$

user
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