Limit of Partial Sums involving Summation of a Product

In summary: I think I've seen something like that before. I'll give it a shot.After some thought, I think I see the idea. We can bound the partial product with a geometric sequence, so we can bound the partial sum with a geometric series, which converges. So the partial sums are bounded, and therefore the sequence of partial sums converges, which implies the original series converges.
  • #1
Euler2718
90
3

Homework Statement


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

Homework Equations

The Attempt at a Solution


[/B]
So we want to find
[tex] \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) [/tex]

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 [itex]\prod_{k=1}^{i}\left( \frac{1}{2} + \frac{1}{k}\right) = \prod_{k=1}^{\infty}\left( 1 + \frac{2-k}{2k}\right)[/itex] the problem then becomes [itex]\sum_{k=1}^{\infty}\frac{2-k}{2k}[/itex] 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?
 
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  • #2
Euler2718 said:

Homework Statement


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

Homework Equations

The Attempt at a Solution


[/B]
So we want to find
[tex] \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) [/tex]

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 [itex]\prod_{k=1}^{i}\left( \frac{1}{2} + \frac{1}{k}\right) = \prod_{k=1}^{\infty}\left( 1 + \frac{2-k}{2k}\right)[/itex] the problem then becomes [itex]\sum_{k=1}^{\infty}\frac{2-k}{2k}[/itex]
Can you show your work leading up to this summation?
Euler2718 said:
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.
Euler2718 said:
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.
 
  • #3
How about calculating the (finite) product first?
 
  • #4
Mark44 said:
Can you show your work leading up to this summation?
The original problem was to prove that [itex]\sum a_{n} [/itex] converges, given that [itex] a_{n+1} = \left( \frac{1}{2} + \frac{1}{n} \right)a_{n} [/itex], with [itex]a_{1}=1[/itex]. After messing around with this, I came up with the formula in my original post.
Mark44 said:
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'.
Mark44 said:
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 said:
How about calculating the (finite) product first?
Alright, I will attempt this.
 
  • #5
Euler2718 said:
The original problem was to prove that [itex]\sum a_{n} [/itex] converges, given that [itex] a_{n+1} = \left( \frac{1}{2} + \frac{1}{n} \right)a_{n} [/itex], with [itex]a_{1}=1[/itex]. 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##.
 
  • #6
Expanding the partial product, I found that [itex]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}}[/itex]. Fortunately this goes to 0 in the limit.
 
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  • #7
Euler2718 said:
Expanding the partial product, I found that [itex]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}}[/itex]. 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##.
 
  • #8
Mark44 said:
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, [itex]\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 [/itex] . I should be looking at the partial sums of this series, correct?
 
  • #9
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?
 
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  • #10
fresh_42 said:
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.
 
  • #11
Ah, so the problem then becomes to show for all [itex]\epsilon>0[/itex] there exists [itex]N\in\mathbb{N}[/itex] such that [itex] \left| \frac{ (n+1)(n+2) }{2^{n-1}} - \frac{ (m+1)(m+2) }{2^{m-1}} \right| < \epsilon [/itex] for all [itex]n,m \geq N[/itex]. Once this is proved, [itex](s_{n})[/itex] is Cauchy, so it must converge, thus the series converges.
 
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  • #12
Euler2718 said:
The original problem was to prove that [itex]\sum a_{n} [/itex] converges, given that [itex] a_{n+1} = \left( \frac{1}{2} + \frac{1}{n} \right)a_{n} [/itex], with [itex]a_{1}=1[/itex]. 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##.
 
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  • #13
Ratio test?
 
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FAQ: Limit of Partial Sums involving Summation of a Product

1. What is the limit of partial sums involving summation of a product?

The limit of partial sums involving summation of a product is a mathematical concept that describes the behavior of a series when the number of terms in that series increases to infinity. It is usually denoted by the symbol "lim" and is used to determine whether a series converges or diverges.

2. How is the limit of partial sums involving summation of a product calculated?

The limit of partial sums involving summation of a product is calculated by taking the sum of all the terms in the series and then taking the limit as the number of terms approaches infinity. This can be done using various mathematical techniques, such as the Ratio Test, the Comparison Test, or the Root Test.

3. What is the significance of the limit of partial sums involving summation of a product?

The limit of partial sums involving summation of a product is important because it helps us determine whether a series is convergent or divergent. This information is essential in many mathematical and scientific applications, such as in the fields of calculus, physics, and statistics.

4. Can the limit of partial sums involving summation of a product be negative?

Yes, the limit of partial sums involving summation of a product can be negative. This depends on the values of the terms in the series and the behavior of the series as the number of terms increases. A series can have a negative limit if the terms in the series alternate between positive and negative values.

5. Are there real-world applications of the limit of partial sums involving summation of a product?

Yes, there are many real-world applications of the limit of partial sums involving summation of a product. For example, it is used in finance to calculate the present value of an investment, in physics to determine the position and velocity of a moving object, and in computer science to analyze the efficiency of algorithms.

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