Evaluate: Σ_(n = 0) ^∞ ((1/(5n + 1)))


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Question Number 75547 by TawaTawa last updated on 12/Dec/19

$Evaluate : \underset{n = 0}{\sum^{\infty}} (\frac{1}{5 n + 1})$
Commented by mind is power last updated on 13/Dec/19

$mor generaly$ $let S = \sum_{k ⩾ 0} \frac{{(- 1)}^{k}}{ak + b}$ $withe (a, b) \in N^{* 2}$ $withe \gcd (a, b) = 1$ $S = \frac{1}{b} . Σ \frac{{(- 1)}^{k}}{1 + \frac{a}{b} k}$ $let f (x) = \frac{1}{1 + x^{\frac{a}{b}}}$ $∣ x ∣< 1 \Rightarrow f (x) = \sum_{k ⩾ 0} {(- 1)}^{k} . x^{\frac{ak}{b}}$ $let ε > 0 \Rightarrow since we have unifirm cv we switch \div$ $\int_{0}^{1 - ε} f (x) dx = \sum_{k ⩾ 0} {(- 1)}^{k} \int_{0}^{1 - ε} x^{\frac{ak}{b}} dx$ $S = \sum_{k ⩾ 0} {(- 1)}^{k} . \frac{1}{1 + \frac{ak}{b}} {(1 - ε)}^{\frac{ak}{b} + 1}$ $we tack ε \to 0 justify by the fact$ $\forall ε > 0 S = \sum_{k ⩾ 0} \frac{{(- 1)}^{k}}{1 + \frac{ak}{b}} {(1 - ε)}^{\frac{ak}{b} + 1} Converge uniformily in compact [0, 1 - ε]$ $\int_{0}^{1} \frac{dx}{1 + x^{\frac{a}{b}}} = S, \frac{a}{b} \neq 1 since \gcd (a, b) = 1$ $x = {tg}^{2 \frac{b}{a}} u$ $\Rightarrow dx = \frac{2 b}{a} (1 + {tg}^{2} (u)) {tg}^{\frac{2 b - a}{a}} (u) du$ $S = \frac{2 b}{a} . \int_{0}^{\frac{π}{4}} {tg}^{\frac{2 b - a}{a}} (u) du$ $S = \frac{2 b}{a} \int_{0}^{\frac{π}{4}} \sin^{\frac{2 b}{a} - 1} (u) . \cos^{\frac{- 2 b}{a} + 2 - 1} (u) du$ $withe$ $\tilde{B} (x, y, s) = 2 \int_{0}^{s} \cos^{2 x - 1} (t) . \sin^{2 y - 1} (t) dt$ $bS = \frac{b}{a} \tilde{B} (\frac{b}{a}, 1 - \frac{b}{a}, \frac{π}{4})$ $\Rightarrow \sum_{k ⩾ 0} \frac{{(- 1)}^{k}}{ak + b} = \frac{\tilde{B} (\frac{b}{a}, 1 - \frac{b}{a}, \frac{π}{4})}{a}, \forall (a, b) \in N^{* 2} withe \gcd (a, b) = 1$ $and (a, b) \neq (1, 1)$
Commented by mathmax by abdo last updated on 12/Dec/19

$p e h a p s t h e Q i s e v a l u a t e \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{5 n + 1}$
Commented by TawaTawa last updated on 12/Dec/19

$ok, help me sir$
Commented by TawaTawa last updated on 13/Dec/19

$God bless you sir$
Commented by mind is power last updated on 13/Dec/19

$y^{'} re Welcom$
	Answered by mind is power last updated on 12/Dec/19

	$diverge + \infty$
	Answered by mind is power last updated on 12/Dec/19
	$Σ_(n≥0) ^(+∞) (1/(5n+1))≥Σ_(n≥2) (1/(5n))=(1/5)Σ_(n≥2) (1/n)......E (1/(n+1))≤∫_n ^(n+1) (1/x)≤(1/n) ⇒Σ_(n=2) ^(+N) (1/n)≥ln(N+1)−ln(2)⇒ lim _(N→∞) Σ_(n=2) ^N (1/n)≥lim_(N→+∞) {ln(N+1)−ln(2)}=+∞ by E we get +∞$
	$\underset{n ⩾ 0}{\sum^{+ \infty}} \frac{1}{5 n + 1} ⩾ \sum_{n ⩾ 2} \frac{1}{5 n} = \frac{1}{5} \sum_{n ⩾ 2} \frac{1}{n} . . . . . . E$ $\frac{1}{n + 1} ⩽ \int_{n}^{n + 1} \frac{1}{x} ⩽ \frac{1}{n}$ $\Rightarrow \underset{n = 2}{\sum^{+ N}} \frac{1}{n} ⩾ \ln (N + 1) - \ln (2) \Rightarrow$ $\lim \underset{N \to \infty}{} \underset{n = 2}{\sum^{N}} \frac{1}{n} ⩾ \lim_{N \to + \infty} {\ln (N + 1) - \ln (2)} = + \infty$ $by E we get + \infty$
	Commented by TawaTawa last updated on 12/Dec/19

	$God bless you sir .$ $Is there shortcut to know a series diverges$
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