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I'm having trouble determining whether the infinite product in $(1.)$ converges or diverges, my attack was to use the convergence cretina stated in $(0.)$

$(0.)$

An Infinite Product is said to be convergent if there exists a non-zero limit of the sequence of partial products:

$$P_{n}=\prod_{k=1}^n(1 + u_{k})$$.

as $n \rightarrow \infty$ The value of the infinite product is the limit: $$P = \lim_{n \rightarrow \infty}P_{n}$$

and one writes $$\prod_{k=1}(1 + u_{k}) = P$$

$(1.)$

$$\prod_{k=2}\left(1 + (-1)^{k}\frac{1}{k}\right)$$

Attacking $(1.)$ via our convergence criteria stated in $(0.)$ one can make the following observations:

$$P_{n} = \prod_{k=2}^n\left(1 + (-1)^{k}\frac{1}{k}\right)$$

Now taking the limit of $P_{n}$ we may now see: $$\lim_{n \rightarrow \infty} \left(\prod_{k=2}^n\left(1 + (-1)^{k}\frac{1}{k}\right)\right)$$

Further observations reveal that we have the following: $$\lim_{n \rightarrow \infty}P_{n}=\lim_{n \rightarrow \infty} \left(\prod_{k=2}^n\left(1 + (-1)^{k}\frac{1}{k}\right)\right)= \lim_{n \rightarrow \infty}\left(P_{2}\times P_{3}\times P_{4}\times P_{5} \times \cdots \times P_{n}\right)$$.

Finally in conclusion the limit on the RHS side of our result becomes the following: $$\lim_{n \rightarrow \infty}\left(1 +(-1)^{2}\frac{1}{2}\right) \times \cdots + \left(1 +(-1)^{n})\frac{1}{n}\right)$$

My question is how to convert our sequence of partial products as seen in our previous result into an infinite series of any form ?

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  • $\begingroup$ Note that, for every $n$, $$P_{2n+1}=\prod_{k=2}^{2n+1}\left(1+\frac{(-1)^k}k\right)=\prod_{i=1}^{n}\left(1+\frac1{2i}\right)\left(1-\frac1{2i+1}\right)=\prod_{i=1}^{n}\frac{2i+1}{2i}\frac{2i}{2i+1}=1$$ and conclude that $$\lim_{n\to\infty}P_n=\ldots$$ $\endgroup$
    – Did
    May 26, 2017 at 12:42
  • $\begingroup$ What you are doing in (1.) is a mystery: sentences with no verb, equalities with no equal sign, and the most mysterious assertion that $$\lim_{n \rightarrow \infty}P_{n}=\lim_{n \rightarrow \infty}P_{2}+P_{3}+P_{4}+P_{5} + \cdot \cdot \cdot + P_{n}$$ Please explain what you mean there. $\endgroup$
    – Did
    May 26, 2017 at 12:45
  • $\begingroup$ And the tag (complex-analysis) is quite off-topic. $\endgroup$
    – Did
    May 26, 2017 at 12:46
  • $\begingroup$ That initial statement was me just taking the limit of partial products $\endgroup$
    – Zophikel
    May 26, 2017 at 12:49
  • $\begingroup$ Well, in this case you should be aware that expanding the partial products does not yield $P_2+P_3+\cdots+P_n$. $\endgroup$
    – Did
    May 26, 2017 at 12:52

3 Answers 3

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You can consider the partial sum of $1)$ even or $2)$ odd number of terms:

$$\color{blue}{1) \lim_\limits{n\to\infty} \prod_{k=2}^{2n}\left(1+\frac{(-1)^k}{k}\right)=}$$ $$\lim_\limits{n\to\infty} \left(1+\frac 12\right)\left(1-\frac 13\right)\left(1+\frac 14\right)\left(1-\frac 15\right)\cdots\left(1+\frac 1{2n-2}\right)\left(1-\frac 1{2n-1}\right)\left(1+\frac 1{2n}\right)=$$ $$\lim_\limits{n\to\infty} \frac 32 \cdot \frac 23\cdot \frac 54\cdot \frac 45\cdots\cdot \frac {2n-1}{2n-2}\cdot\frac {2n-2}{2n-1}\cdot\frac {2n+1}{2n}=\lim_\limits{n\to\infty} \frac {2n+1}{2n}=1.$$

$$\color{blue}{2) \lim_\limits{n\to\infty} \prod_{k=2}^{2n+1}\left(1+\frac{(-1)^k}{k}\right)}=$$ $$\lim_\limits{n\to\infty} \left(1+\frac 12\right)\left(1-\frac 13\right)\left(1+\frac 14\right)\left(1-\frac 15\right)\cdots\left(1+\frac 1{2n}\right)\left(1-\frac 1{2n+1}\right)=$$ $$\lim_\limits{n\to\infty} \frac 32 \cdot \frac 23\cdot \frac 54\cdot \frac 45\cdots\cdot\frac {2n+1}{2n}\cdot\frac {2n}{2n+1}=\lim_\limits{n\to\infty} 1=1.$$

Note: It is possible to take logarithm to convert the infinite product to a series.

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Hint. Consider the sequence of partial products: $$P_N=\prod_{n=2}^N\left(1+(-1)^{n}\frac{1}{n}\right).$$ They have a closed formula.

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  • $\begingroup$ Interesting I didn't think of this at the time $\endgroup$
    – Zophikel
    May 26, 2017 at 12:42
  • $\begingroup$ Why add the last sentence? The post was better without it... $\endgroup$
    – Did
    May 26, 2017 at 12:43
  • $\begingroup$ @Did OK, I did not see your comment above. $\endgroup$
    – Robert Z
    May 26, 2017 at 12:45
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Hint:

For the odd partial products: $$ \begin{align} \prod_{n=2}^{2m+1}\left(1+\frac{(-1)^n}n\right) &=\prod_{k=1}^m\left(1+\frac1{2k}\right)\left(1-\frac1{2k+1}\right) \end{align} $$ For the even partial products $$ \begin{align} \prod_{n=2}^{2m}\left(1+\frac{(-1)^n}n\right) &=\left(1+\frac1{2m}\right)\prod_{k=1}^{m-1}\left(1+\frac1{2k}\right)\left(1-\frac1{2k+1}\right) \end{align} $$

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