Find out the answer to "How do I find: ∏n=1∞(1+1π2n2)"

pozicijombx

Answered question

2022-01-23

How do I find:

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

nebajcioz

Beginner2022-01-24Added 15 answers

The product converges because

\sum n^{- 2}

does. Can you see why? What is the greatest sum that can possibly appear when we expand the product?
In fact, suppose that

a_{n} \geq 0

for each n. Set

p_{n} = \prod_{k = 1}^{n} a_{k}

Then

{\log p}_{n} = \sum_{k = 1}^{n} {\log a}_{k}

\sum a_{k}

converges, then

a_{k} \to 0

, then since

lim_{x \to 0} \frac{\log (1 + x)}{x} = 1

so that

lim \frac{\log (1 + a_{n})}{a_{n}} = 1

the comparison test tells us

{\log p}_{n}

converges, say to l. By continuity of the logarithm,

p_{n}

must converge to p with

\log p = l

so that

lim p_{n} = e^{l}

Conversely, suppose

{\log p}_{n} = \sum_{k = 1}^{n} {\log a}_{k}

converges. This means that

\log (1 + a_{k}) \to 0

so that

a_{k} \to 0

. Comparison yields from

lim \frac{\log (1 + a_{n})}{a_{n}} = 1

that

\sum a_{k}

converges too. Thus, we have shown

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