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We know that the odd terms of Dirichelt Beta function are $\frac{\pi }{4}$,$\frac{\pi^3 }{32}$,$\frac{5\pi^5 }{1536}$,... If we use the WolframAlpha to find the limit of the odd terms only divided by $n$ when the $N\rightarrow \infty $ as shown in below equation

$$\lim_{N\rightarrow \infty }\sum_{n=1}^{N }\frac{\beta (2n-1)}{n}$$ The results of WolframAlpha were as shown below enter image description here enter image description here enter image description here we see the limit go to $1$

But if we take the limit for all terms (add and even) as shown below $$\lim_{N\rightarrow \infty }\sum_{n=1}^{N }\frac{\beta (n)}{n}$$ the result become

enter image description here enter image description here enter image description here

I cannot continue for more $n$ because the ability of Wolfarm stopped up to $n=550$

We see the limit also go to $1$

My question is "Which limit will give $1$"

E.H.E
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  • You cannot take the limit of a summation variable, i.e. $\lim_{n\to \infty} \sum_{n=0}^\infty$ does not make sense. Did you perhaps mean $\lim_{N\to \infty} \sum_{n=0}^N$ ? – Winther Dec 28 '14 at 11:03
  • Since the denominator is $n$ in both expressions, they could both converge to $1$. – Alice Ryhl Dec 28 '14 at 11:58
  • $\displaystyle\sum_{n=1}^\infty\frac{\beta(2n-1)}n ~=~ \eta(1)-\frac{\beta(2)}{\beta(1)} ~=~ \ln2-4\frac{\text{Catalan}}\pi ~\simeq~ -0.473\ldots$ – Lucian Dec 28 '14 at 15:33

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