Converting an $\prod_{n=2}(1+(-1)^{n}\frac{1}{n})$ into an Infinite Series to determine Convergence?

Question

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 ?

Answer 1

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.

Answer 2

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.

Answer 3

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 $$ \begin{align} \prod_{n=2}^{2m}\left(1+\frac{(-1)^n}n\right) &=\left(1+\frac1{2m}\right)\prod_{k=1}^{m-1}\left(1+\frac1{2k}\right)\left(1-\frac1{2k+1}\right) \end{align} $$