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$$\sum_{n=1}^\infty \left(1+\frac{1}{2}+\dots+\frac{1}{n}\right)x^n$$

I think we can write it in this form also: $\sum_{n=1}^\infty \left(\sum_{k=1}^{n}\frac{1}{k}\right)x^n$

I tried the ratio test: $$\lim_{n\to\infty} \left|\frac{\left(\sum_{k=1}^{n+1}\frac{1}{k}\right)x^{n+1}}{\left(\sum_{k=1}^{n}\frac{1}{k}\right)x^{n}}\right|$$

Here, I don't know what to do with those summations, because they're both infinity.

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    What have you tried? Hint: ratio test. – Sean Roberson Jan 06 '23 at 23:16
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    I tried ratio test but, I becomes to an infinity divided by infinity. I don't know how to face it here. –  Jan 06 '23 at 23:18
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    Please show us, in full, that attempt, even if it fails. Edit your original post to include this. – Sean Roberson Jan 06 '23 at 23:18
  • These are some ideas that all solve this problem. (1) Note that the coefficients are partial sums. and the factors $x^n$ are easy to sum. Therefore, the series becomes "simpler" if you apply summation by parts – plop Jan 06 '23 at 23:24
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    I don't know how to use summation by parts here. I've never used them. Can you explain more please? –  Jan 06 '23 at 23:38
  • $\sum_{k=1}^n\frac1k$$\sim\ln(n)$ can help you for the ratio test. – Anne Bauval Jan 06 '23 at 23:40
  • This is equivalent to the application of summation by part, and perhaps easier for you to see. Assume that $x$ is a value for which the series converges absolutely and call $S(x)$ the sum. Then compute $(1-x)S(x)=S(x)-xS(x)$. Note that $xS(x)$ is almost the same series but shifted. Cancel from $S(x)$ and $-xS(x)$ many many terms that have the same coefficient and the same power of $x$. You should get $(1-x)S(x)=\sum_{n=1}^{\infty}\frac{x^n}{n}$. – plop Jan 06 '23 at 23:45
  • Harmonic series generating function – user170231 Jan 07 '23 at 00:17
  • @user170231 That link was already present at the end of my answer before seeing your comment, but just as a complement for future readers, because the present question is more elementary. – Anne Bauval Jan 07 '23 at 08:19

2 Answers2

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If $x\in [1,\infty),$ then $\sum_{n=1}^\infty \left(1+\frac{1}{2}+\dots+\frac{1}{n}\right)x^n\ge \sum_{n=1}^\infty x^n$ so the series diverges there.

So suppose $0\le x<1.$ Then, since $2+\frac{1}{3}+\dots+\frac{1}{n+1}\le\int_1^{n+1}\frac{1}{t}dt,$ we have

$\sum_{n=1}^\infty \left(1+\frac{1}{2}+\dots+\frac{1}{n}\right)x^n\le x^n-\frac{x^{n+1}}{n+1}+x^n\ln (n+1)$

so we conclude that the series converges (the last term converges, for example, by L'Hospital's rule.)

Now we are done, because the series has an interval of convergence symmetric about $x=0,$ so the foregoing implies that it converges in $(-1,1).$

Matematleta
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For any integer $n\ge1,$ $$H_n:=\sum_{k=1}^n\frac1k\ge1$$ so, as $n\to+\infty,$ $$\frac{H_{n+1}}{H_n}=1+\frac1{(n+1)H_n}\to1$$ and $$\frac{H_{n+1}|x|^{n+1}}{H_n|x|^n}\to|x|$$ hence the radius of convergence of this power series is equal to $1.$

Therefore, the series is absolutely convergent when $|x|<1,$ and trivially divergent (i.e. its general term does not tend to $0$) when $|x|>1.$

When $x=\pm1,$ it is also trivially divergent.

See the sum of this power series.

Anne Bauval
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