Limit of Partial Sums involving Summation of a Product

[Euler2718]

Homework Statement

Show that the sequence of partial sums
$$s_{n} = 1+\sum_{i=1}^{n} \left(\prod_{k=1}^{i}\left( \frac{1}{2} + \frac{1}{k}\right)\right)$$
converges, with $n\in \mathbb{N}\cup \{0\}$

Homework Equations

The Attempt at a Solution

[/B]
So we want to find
$$\lim_{n\to\infty} s_{n} = \lim_{n\to\infty} \left(1+ \sum_{i=1}^{n} \left(\prod_{k=1}^{i}\left( \frac{1}{2} + \frac{1}{k}\right)\right) \right) = 1 + \sum_{i=1}^{\infty} \left(\prod_{k=1}^{i}\left( \frac{1}{2} + \frac{1}{k}\right) \right)$$

I haven't had much experience in a summation of a product, so my best guess was to determine convergence of the product first. Rewriting $\prod_{k=1}^{i}\left( \frac{1}{2} + \frac{1}{k}\right) = \prod_{k=1}^{\infty}\left( 1 + \frac{2-k}{2k}\right)$ the problem then becomes $\sum_{k=1}^{\infty}\frac{2-k}{2k}$ which diverges by the limit comparison test. If I recall correctly this implies the product should diverge to 0? So the infinite sum of this is just 0, and the limit of partial sums is then 1? However wolfram alpha begs to differ and it gives me the answer of 8. How should I have proceeded?

 

[Mark44]

Homework Statement

Show that the sequence of partial sums
$$s_{n} = 1+\sum_{i=1}^{n} \left(\prod_{k=1}^{i}\left( \frac{1}{2} + \frac{1}{k}\right)\right)$$
converges, with $n\in \mathbb{N}\cup \{0\}$

Homework Equations

The Attempt at a Solution

[/B]
So we want to find
$$\lim_{n\to\infty} s_{n} = \lim_{n\to\infty} \left(1+ \sum_{i=1}^{n} \left(\prod_{k=1}^{i}\left( \frac{1}{2} + \frac{1}{k}\right)\right) \right) = 1 + \sum_{i=1}^{\infty} \left(\prod_{k=1}^{i}\left( \frac{1}{2} + \frac{1}{k}\right) \right)$$

I haven't had much experience in a summation of a product, so my best guess was to determine convergence of the product first. Rewriting $\prod_{k=1}^{i}\left( \frac{1}{2} + \frac{1}{k}\right) = \prod_{k=1}^{\infty}\left( 1 + \frac{2-k}{2k}\right)$ the problem then becomes $\sum_{k=1}^{\infty}\frac{2-k}{2k}$

Can you show your work leading up to this summation?

which diverges by the limit comparison test. If I recall correctly this implies the product should diverge to 0?

What does "diverge to 0" mean? You could say that a product converges to some number, but you can't say that it diverges to a specific number.

So the infinite sum of this is just 0, and the limit of partial sums is then 1? However wolfram alpha begs to differ and it gives me the answer of 8. How should I have proceeded?

I would try to expand the sum and product for a few terms to get an idea of possible simplification, or at least a general idea of the behavior.

 

[fresh_42]

How about calculating the (finite) product first?

 

[Euler2718]

Can you show your work leading up to this summation?

The original problem was to prove that $\sum a_{n}$ converges, given that $a_{n+1} = \left( \frac{1}{2} + \frac{1}{n} \right)a_{n}$, with $a_{1}=1$. After messing around with this, I came up with the formula in my original post.

What does "diverge to 0" mean? You could say that a product converges to some number, but you can't say that it diverges to a specific number.

Sorry, I was referring to jargon used here
https://math.stackexchange.com/questions/778202/convergence-of-infinite-product-prod-n-2-infty-1-frac-1n
Would probably be better to stick to 'converges'.

I would try to expand the sum and product for a few terms to get an idea of possible simplification, or at least a general idea of the behavior.

How about calculating the (finite) product first?

Alright, I will attempt this.

 

[Mark44]

The original problem was to prove that $\sum a_{n}$ converges, given that $a_{n+1} = \left( \frac{1}{2} + \frac{1}{n} \right)a_{n}$, with $a_{1}=1$. After messing around with this, I came up with the formula in my original post.

After working with this formula for a few terms in the sequence, I think I have come up with a formula for $a_{n + 1}$ in terms of $a_0$.

 

[Euler2718]

Expanding the partial product, I found that $p_{n} = \left(\frac{1}{2} + 1\right)\cdot\left(\frac{1}{2} + \frac{1}{2}\right) \dots \cdot \frac{(n+1)(n+2)}{2^{n-1}}$. Fortunately this goes to 0 in the limit.

 

[Mark44]

Expanding the partial product, I found that $p_{n} = \left(\frac{1}{2} + 1\right)\cdot\left(\frac{1}{2} + \frac{1}{2}\right) \dots \cdot \frac{(n+1)(n+2)}{2^{n-1}}$. Fortunately this goes to 0 in the limit.

Which is irrelevant in your problem.

The fact that $\lim_{n \to \infty}a_n = 0$ doesn't have any bearing on whether $\sum_{n = 0}^\infty a_n$ converges or diverges. For example, $\sum_{n = 1}^\infty \frac 1 n$ diverges, while $\sum_{n = 1}^\infty \frac 1 {n^2}$ converges. In both cases, $\lim_{n \to \infty}a_n = 0$.

 

[Euler2718]

Which is irrelevant in your problem.

My original attack on the problem was to analyze the product first to see it it converged. I haven't been able to make a conclusion about the series yet. I know that the problem I should be looking at is, $\sum_{i=1}^{\infty}\left( \prod_{k=1}^{i} \left( \frac{1}{2} + \frac{1}{k} \right) \right) =\prod_{k=1}^{1} \left( \frac{1}{2} + \frac{1}{k} \right) +\prod_{k=1}^{2} \left( \frac{1}{2} + \frac{1}{k} \right) +\prod_{k=1}^{3} \left( \frac{1}{2} + \frac{1}{k} \right) + \dots$ . I should be looking at the partial sums of this series, correct?

 

[fresh_42]

You have calculated the product, I think. At least I found the correct term in what you've written. I just didn't know what $p_n$ should be. Now you can calculate $|s_n-s_m|$. Does this sound familiar?

 

[Euler2718]

You have calculated the product, I think. At least I found the correct term in what you've written. I just didn't know what $p_n$ should be. Now you can calculate $|s_n-s_m|$. Does this sound familiar?

If I'm reading it right, that difference is a set up for Cauchy criteria.

 

[Euler2718]

Ah, so the problem then becomes to show for all $\epsilon>0$ there exists $N\in\mathbb{N}$ such that $\left| \frac{ (n+1)(n+2) }{2^{n-1}} - \frac{ (m+1)(m+2) }{2^{m-1}} \right| < \epsilon$ for all $n,m \geq N$. Once this is proved, $(s_{n})$ is Cauchy, so it must converge, thus the series converges.

 

[Ray Vickson]

The original problem was to prove that $\sum a_{n}$ converges, given that $a_{n+1} = \left( \frac{1}{2} + \frac{1}{n} \right)a_{n}$, with $a_{1}=1$. After messing around with this, I came up with the formula in my original post.

Sorry, I was referring to jargon used here
https://math.stackexchange.com/questions/778202/convergence-of-infinite-product-prod-n-2-infty-1-frac-1n
Would probably be better to stick to 'converges'.Alright, I will attempt this.

If all you want to do is show convergence of $\sum a_n$ then you do not need such fancy machinery; perhaps you need to do much more if you want to actually evaluate the summation. Beyond a certain finite value of $n$ the $a_n$ are bounded above by a geometric sequence $c \alpha^n$, with $0 < \alpha < 1$.

 

[hunt_mat]

Ratio test?