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What is the limit of

$$\sum_{n=1}^{m} \frac{n}{1.3.5...(2n+1)}$$

when $m \to \infty$ ?

I tried to find a recursive formula but failed to put things on place ?

Also I had a little doubt that when we say a series is converging then does we mean that terms of our series doesn't explode to infinite or sum of the series doesn't go to infinite ?

Jean Marie
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RKK
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  • This has been asked and answered before: https://math.stackexchange.com/q/2257306/42969, https://math.stackexchange.com/q/3090048/42969 – Martin R Sep 17 '21 at 14:33
  • Render the term as $(1/2)[1/(1×3×5×...×(2n-1))-1/(1×3×5×...×(2n+1))]$ and telescope. – Oscar Lanzi Sep 17 '21 at 14:51
  • There is a specific notation for your denominators which is the double factorial – Jean Marie Sep 17 '21 at 15:38
  • @Oscar Lanzi Okay thanks I got it . Can you see My little extra doubt ? – RKK Sep 19 '21 at 05:21
  • Converging means there is a definite limit, not just the sum is bounded. The series $1-1+1-1+1-1+...$ remains bounded but can't decide between $1$ and $0$, alternating between these sums, so we call it divergent. Your series, however, really does converge. We just saw the proof. – Oscar Lanzi Sep 19 '21 at 10:17

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