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If $x >0$, consider the series $\sum\limits_{n=1}^\infty a_n$, where $$a_{n}=(2-x)(2-x^{1/2})(2-x^{1/3})\cdots(2-x^{1/n}).$$ Is it convergent?

This question is in my textbook. I don't know how to solve it. When $0<x\le1$, the series is not convergent since $a_{n}\ge1$. But how to think about the situations when $x >1$?

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  • when 0<x<=1, the series is divergent since $a_{n}$>=1, but I am wondering how to analyze the situation when x>1? – python3 Jul 22 '14 at 06:48

2 Answers2

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If $x\leqslant1$, then $a_n\geqslant1$ for every $n$ hence the series $\sum\limits_na_n$ diverges.

If $x\gt1$, then the expansion $x^{1/n}=\exp((\log x)/n)=1+(\log x)/n+o(1/n)$ yields $$\log(a_{n+1}/a_n)=\log(2-x^{1/n})\sim-(\log x)/n,$$ hence $\log a_n\sim-(\log x)\cdot\log n$. Thus:

  • If $x\gt\mathrm e$, one can pick $1\lt y\lt\log x$ then $a_n\leqslant1/n^y$ for every $n$ large enough, which proves that the series $\sum\limits_na_n$ converges.
  • If $x\lt\mathrm e$, then $a_n\geqslant1/n$ for every $n$ large enough, which proves that the series $\sum\limits_na_n$ diverges.

If $x=\mathrm e$, more care is required. Note that in this case, $$\log(a_{n+1}/a_n)=\log(2-\mathrm e^{1/n})=\log(1-1/n+O(1/n^2))=-1/n+O(1/n^2),$$ hence $\log a_n=-\log n+O(1)$, in particular, $a_n\geqslant C/n$ for every $n$ large enough, for some positive $C$, hence the series $\sum\limits_na_n$ diverges.

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you say this appears in your textbook. It is only wise to learn all convergence criteria in the corresponding chapter or section, for the exercises in the end are based thereon.

Raabe's convergence criterion applies for this series.

Here's a nice exposé. http://en.wikipedia.org/wiki/Ratio_test

One need calculate, for a series $\sum a_n$ (here $a_n \geq 0$), the limit $\lambda=lim\ n(\frac{a_n}{a_{n+1}}-1).$

(There are more refined versions with lim sup, lim inf, but this will do.)

If $\lambda < 1$, the series is divergent. The series is convergent for $\lambda>1$ and inconclusive for $\lambda=1.$ One way to prove this criterion is by comparison with a suitable harmonic series (compare with the above solution), but this is not the only proof that exists.

For decreasing series, it is good to work out an asymptotic formula for the quotient (see the above link). You should get some basic reference and collect convergence criteria, too. Hope this helps.

Thus, in this case we have the function $n(x^{1/n}-1)$ and, if you don't know the Taylor formula or can't apply it yet, it suffices to know derivatives to calculate:

$$lim_{h \to 0} \frac{a^h-1}{h},$$ and to substitute $h=1/n.$

Here's an internal link to Raabe's test. The proof of Raabe's Test for absolute convergence

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