I need to find the limit of the following sequence: limn→∞∏k=1n(1+kn2)

Note that we have log⁡(∏k=1n(1+kn2))=∑k=1nlog⁡(1+kn2) (1) Applying the right-hand side inequality in (1) reveals ∑k=1nlog⁡(1+kn2)≤∑k=1nkn2 =n(n+1)2n2 =12+12n (2) Applying the left-hand side inequality in (1) reveals ∑k=1nlog⁡(1+kn2)≥∑k=1nkk+n2 ≥∑k=1nkn+n2 =n(n+1)2(n2+n) =12 (3) Putting 1-3 together yields 12≤log⁡(∏k=1n(1+kn2))≤12+12n (4) whereby application of the squeeze theorem to (4) gives limn→∞log⁡(∏k=1n(1+kn2))=12 Hence, we find that limn→∞∏k=1n(1+kn2)=e

Note that∏k≤n(1+kn2)=∏k≤n(n2+kn2)=(n2+1)nn2nwhere (x)m is the Pochhammer symbol and since(x)m=Γ(x+m)Γ(x)we have∏k≤n(1+kn2)=Γ(n2+n+1)n2nΓ(n2+1)→ewhere the last limit can be calculated using the Stirling's approximation.

Hint Pn=∏k=1n(1+kn2)⇒log⁡(Pn)=∑k=1nlog⁡(1+kn2) Work with the sum (without forgetting by the end that Pn=elog⁡(Pn))

Note that we have log⁡(∏k=1n(1+kn2))=∑k=1nlog⁡(1+kn2) (1) Applying the right-hand side inequality in (1) reveals ∑k=1nlog⁡(1+kn2)≤∑k=1nkn2 =n(n+1)2n2 =12+12n (2) Applying the left-hand side inequality in (1) reveals ∑k=1nlog⁡(1+kn2)≥∑k=1nkk+n2 ≥∑k=1nkn+n2 =n(n+1)2(n2+n) =12 (3) Putting 1-3 together yields 12≤log⁡(∏k=1n(1+kn2))≤12+12n (4) whereby application of the squeeze theorem to (4) gives limn→∞log⁡(∏k=1n(1+kn2))=12 Hence, we find that limn→∞∏k=1n(1+kn2)=e

Note that∏k≤n(1+kn2)=∏k≤n(n2+kn2)=(n2+1)nn2nwhere (x)m is the Pochhammer symbol and since(x)m=Γ(x+m)Γ(x)we have∏k≤n(1+kn2)=Γ(n2+n+1)n2nΓ(n2+1)→ewhere the last limit can be calculated using the Stirling's approximation.

Hint Pn=∏k=1n(1+kn2)⇒log⁡(Pn)=∑k=1nlog⁡(1+kn2) Work with the sum (without forgetting by the end that Pn=elog⁡(Pn))

Find out the answer to "I need to find the limit of the..."

Jazmin Perry

Answered question

2022-01-24

I need to find the limit of the following sequence:

lim_{n \to \infty} \prod_{k = 1}^{n} (1 + \frac{k}{n^{2}})

Answer & Explanation

bemolizisqt

Beginner2022-01-25Added 16 answers

Note that we have

\log (\prod_{k = 1}^{n} (1 + \frac{k}{n^{2}})) = \sum_{k = 1}^{n} \log (1 + \frac{k}{n^{2}})

(1)
Applying the right-hand side inequality in (1) reveals

\sum_{k = 1}^{n} \log (1 + \frac{k}{n^{2}}) \leq \sum_{k = 1}^{n} \frac{k}{n^{2}}

= \frac{n (n + 1)}{2 n^{2}}

= \frac{1}{2} + \frac{1}{2 n}

(2)
Applying the left-hand side inequality in (1) reveals

\sum_{k = 1}^{n} \log (1 + \frac{k}{n^{2}}) \geq \sum_{k = 1}^{n} \frac{k}{k + n^{2}}

\geq \sum_{k} = 1^{n} \frac{k}{n + n^{2}}

= \frac{n (n + 1)}{2 (n^{2} + n)}

= \frac{1}{2}

(3)
Putting 1-3 together yields

\frac{1}{2} \leq \log (\prod_{k = 1}^{n} (1 + \frac{k}{n^{2}})) \leq \frac{1}{2} + \frac{1}{2 n}

(4)
whereby application of the squeeze theorem to (4) gives

lim_{n \to \infty} \log (\prod_{k = 1}^{n} (1 + \frac{k}{n^{2}})) = \frac{1}{2}

Hence, we find that

lim_{n \to \infty} \prod_{k = 1}^{n} (1 + \frac{k}{n^{2}}) = \sqrt{e}

Armani Dyer

Beginner2022-01-26Added 10 answers

Note that

\prod_{k \leq n} (1 + \frac{k}{n^{2}}) = \prod_{k \leq n} (\frac{n^{2} + k}{n^{2}}) = \frac{{(n^{2} + 1)}_{n}}{n^{2 n}}

where

{(x)}_{m}

is the Pochhammer symbol and since

{(x)}_{m} = \frac{Γ (x + m)}{Γ (x)}

we have

\prod_{k \leq n} (1 + \frac{k}{n^{2}}) = \frac{Γ (n^{2} + n + 1)}{n^{2 n} Γ (n^{2} + 1)} \to \sqrt{e}

where the last limit can be calculated using the Stirling's approximation.

RizerMix

Expert2022-01-27Added 656 answers

Hint

P_{n} = \prod_{k = 1}^{n} (1 + \frac{k}{n^{2}}) \Rightarrow \log (P_{n}) = \sum_{k = 1}^{n} \log (1 + \frac{k}{n^{2}})

Work with the sum (without forgetting by the end that

P_{n} = e^{\log (P_{n})}

)

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