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Question is like in the title and my attempt is

Let have sequence $$a_n = <\frac{H_0}{10^0},\frac{H_1}{10^1},\frac{H_2}{10^2},\dots>$$ where $H_n$ is n-th harmonic number.

And we have to construct $$b_n = \langle a_0,a_0+a_1,a_0+a_1+a_2,\dots\rangle$$ and it will be our answer.

Let's have generating function for sequence $a_n$: $$A(x) = \sum_{n=0}^\infty a_n x^n$$

And generating function for sequence $b_n$ will be $$B(x) = \frac{A(x)}{1-x}$$

I started making $\frac{H_n}{10^n}$ a bit easier(there are my doubts).

we can rewrite $H_n$ as $$1+\dfrac12+\dfrac13+\dots = 1+\frac{\frac{n(n+1)}{2}-\dfrac22}{n!} = 1+\frac{n(n+1)-2}{2*n!}$$ so now we can rewrite $a_n$ as $$a_n =\langle\frac{1}{10^n}+\frac{n(n+1)-2}{2*n!*10^n}\rangle$$

Generating function for $$\sum_{n=0}^\infty \frac{1}{10^n}x^n = \frac{1}{1-\frac{1}{10}x}$$

But i'm having problem with second part of it... And i'm not sure if i don't overcomplicated this task by making it "easier". I'd like to use generating functions cause it's task for them, but i'd be happy to see some other solutions. Thanks in advance.

Krzysztof Lewko
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  • Please consider \langle and \rangle to give $\langle$ and $\rangle$ instead of $<$ and $>$. Moreover, please consider \left\langle \right\rangle in pairs to give large pairs: $$\left\langle \frac{a}{b}\right\rangle$$ – Fly by Night Mar 27 '14 at 17:47
  • The summation of the $H_n$ you did is not correct. – OR. Mar 27 '14 at 17:51
  • You have a sequence that is easy to differentiate $\Delta H_n=1/(n+1)$ and a sequence (geometric) that is easy to integrate $\sum_{n=1}^{m}1/10^n=\frac{10^{-m-1}-1}{10^{-1}-1}$. So, summation by parts. – OR. Mar 27 '14 at 17:57

3 Answers3

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It is known that

$$\sum_{k=1}^{\infty} H_n x^n = -\frac{\log{(1-x)}}{1-x}$$

Plug in $x=1/10$.

I don't think $H_0$ is defined.

EDIT

You can define $H_0=0$ using the expression

$$H_n = \int_0^1 dx \frac{1-x^n}{1-x}$$

Ron Gordon
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1

You can use the well known identity

$$\sum _{n=0}^{\infty } H_{n}{z}^{n}= \frac{\ln(1-z)}{z-1}. $$

Mhenni Benghorbal
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    Oh wow... haven't thought about it this way! Amazing way of thinking, thanks! – Krzysztof Lewko Mar 27 '14 at 17:58
  • @Chris: You are very welcome. – Mhenni Benghorbal Mar 27 '14 at 17:59
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    I want to be sure, so sorry about novice question. If we want to calculate anything from $$\sum_{n=0}^\infty a_nx^n$$ we need to be sure that this sum converges right? i mean $$\sum x^n = \frac{1}{1-x}$$, but to plug in some $x$ we have to know that this sum converges for given $x$? – Krzysztof Lewko Mar 27 '14 at 18:03
  • And in general, from a generating function $f$ of a sequence, to get the generating function of the partial sums, you can try $f(x)/(1-x)$. – GEdgar Mar 27 '14 at 18:04
  • @Chris: Generally, getting a closed form for the ordinary generating function depends on $a_n$. see for example here. – Mhenni Benghorbal Mar 27 '14 at 18:07
  • @MhenniBenghorbal, yes i know that, but let's say we have to calculate $$\sum_{n=0}^\infty \dfrac12$$ I think we can plug in $x=\dfrac12$ into $$\sum x^n = \frac{1}{1-x}$$, because $\frac{1}{2^n}\rightarrow 0$? Or my logic is wrong? – Krzysztof Lewko Mar 27 '14 at 18:16
  • @Chris: Are you trying to derive some conditions on $a_n$ so you can ensure the convergence? – Mhenni Benghorbal Mar 27 '14 at 18:23
  • @moderators: As I said before "the option of downvoting has been abused by some users". It is the second downvote on this post!! – Mhenni Benghorbal Mar 27 '14 at 18:24
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It's straightforward to prove the identity by formal manipulation actually. You want $\displaystyle\sum_{n \ge 0} H_n \frac{1}{10^n}$, so let's consider the generating function $A(z) = \sum_{n \ge 0} H_n z^n$, and evaluating $A(\frac1{10})$ will give the answer you want.

By definition, we have $$A(z) = \sum_{n \ge 0} H_n z^n = \sum_{n \ge 0} \sum_{k=1}^{n} \frac{1}{k} z^n = \sum_{k \ge 1} \frac1k \sum_{n \ge k} z^n = \sum_{k \ge 1} \frac1k \frac{z^k}{1-z} = \frac{1}{1-z} \sum_{k \ge 1} \frac{z^k}{k} = \frac{1}{1-z} \ln \frac{1}{1-z}.$$

The last step, the well-known series for $\displaystyle \ln \frac{1}{1-z}$, can itself be derived as follows (interpret $\int (\cdot)$ as $\int_0^z (\cdot)\, dz$):

$$\sum_{k \ge 1} \frac{z^k}{k} = \sum_{k \ge 0} \int z^k = \int \sum_{k \ge 0} z^k = \int \frac{1}{1-z} = -\ln(1-z).$$

ShreevatsaR
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