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Evaluate the convergence of the following infinite sum. I am so bad when it comes to parameters. I don't know where to start. $\sum_{n=1}^{\infty}\dfrac{(n+1)^a-n^a}{n^b}$, where $a,b\in\mathbb{R}$.

PaulDirac
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2 Answers2

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Hint. As $n \to \infty$, by using a Taylor series expansion, one may write $$ \left(1+\frac1{n}\right)^a=1+\frac{a}n+O\left(\frac{1}{n^{2}}\right) $$ giving $$ \frac{(n+1)^a-n^a}{n^b}=\frac{1}{n^{b-a}}\cdot\left(\left(1+\frac1{n}\right)^a-1\right)=\frac{a}{n^{b-a+1}}+O\left(\frac{1}{n^{b-a+2}} \right). $$

Olivier Oloa
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  • Your solutions sounds awesome, but sadly enough,I haven't learned Taylor series yet. I'll attend a lecture in some weeks. Till then, is there any solution without it? – PaulDirac Nov 21 '16 at 14:59
  • Thanks. One may use some standard inequalities instead. Do you know Bernoulli's inequality? – Olivier Oloa Nov 21 '16 at 15:08
  • Yes I do. But I wouldn't have thought about it! I will try to figure it out – PaulDirac Nov 21 '16 at 15:28
  • I've read some stuff about power series and function. And I understood what is a Taylor series (also Maclaurin). I've seen some methods on how to find the Taylor series of a function and so on and so forth. Could you please be just a bit more specific about what $O\left(\dfrac{1}{n^2}\right)$ is? – PaulDirac Nov 21 '16 at 21:13
  • Sure, as $n \to \infty$, $u_n=O\left(\dfrac{1}{n^2}\right)$ means there exists a constant $C$ such that $|u_n|\le \dfrac{C}{n^2}$ as $n \to \infty$. – Olivier Oloa Nov 21 '16 at 21:15
  • @DenisNichita $(n+1)^a-n^a = \int_n^{n+1} a x^{a-1}dx =a n^{a-1}+a\int_n^{n+1} (x^{a-1}-n^{a-1})dx$ $=a n^{a-1}+a\int_n^{n+1} \int_n^x (a-1)t^{a-2}dtdx$ i.e. $|(n+1)^a-n^a- a n^{a-1}| = |a(a-1)\int_n^{n+1} \int_n^x t^{a-2}dtdx|$ $ \le|a(a-1)|\int_n^{n+1} \int_n^x |t^{a-2}|dtdx \le |a(a-1)|\int_n^{n+1} \int_n^x |(n+1)^{a-2}|dtdx = |a(a-1)|\frac{|(n+1)^{a-2}|}{2}$ – reuns Nov 21 '16 at 22:02
  • @user1952009 Very interesting what you're saying. And new to me. Thx! – PaulDirac Nov 22 '16 at 07:40
  • @DenisNichita it basically describes the asymptotic behavior of the function. – Simply Beautiful Art Nov 22 '16 at 11:52
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According to binomial expansion, we have

$$(n+1)^a=n^a+an^{a-1}+\frac{a(a-1)}2n^{a-2}+\dots$$

So, we have

$$\frac{(n+1)^a-n^a}{n^b}=an^{a-1-b}+\frac{a(a-1)}2n^{a-2-b}+\dots$$

Which is equivalent to Oliver's answer.

Summating this for $n\ge1$, we get

$$=a\zeta(1+b-a)+\frac{a(a-1)}2\zeta(2+b-a)+\dots$$

Assuming we have $b>a$ so that it converges. For $a\in\mathbb N$, we get a closed form solution in terms of the zeta function:

$$\sum_{n\ge1}{(n+1)^a-n^a\over n^b}=a\zeta(1+b-a)+\frac{a(a-1)}2\zeta(2+b-a)+\dots+\frac{a(a-1)(a-2)\dots3\cdot2\cdot1}{1\cdot2\cdot3\cdot\ldots(a-1)a}\zeta(b)$$

Though I can't expect much better out of this.

The special case is $a=0$, whereupon the whole sum is $0$.

Simply Beautiful Art
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