Evaluate-n-0-1-5n-1-

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

\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} \dots \dots 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