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




Question Number 75547 by TawaTawa last updated on 12/Dec/19
      Evaluate:     Σ_(n = 0) ^∞ ((1/(5n + 1)))
$$\:\:\:\:\:\:\mathrm{Evaluate}:\:\:\:\:\:\underset{\mathrm{n}\:=\:\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{5n}\:+\:\mathrm{1}}\right) \\ $$
Commented by mind is power last updated on 13/Dec/19
mor generaly  let S=Σ_(k≥0) (((−1)^k )/(ak+b))  withe  (a,b)∈N^(∗2)   withe gcd(a,b)=1  S=(1/b).Σ(((−1)^k )/(1+(a/b)k))  let f(x)=(1/(1+x^(a/b) ))  ∣x∣<1⇒f(x)=Σ_(k≥0) (−1)^k .x^((ak)/b)   let ε>0⇒  since we have unifirm cv we switch ÷  ∫_0 ^(1−ε) f(x)dx=Σ_(k≥0) (−1)^k ∫_0 ^(1−ε) x^((ak)/b) dx  S=Σ_(k≥0) (−1)^k .(1/(1+((ak)/b)))(1−ε)^(((ak)/b)+1)   we tack ε→0  justify by the fact  ∀ε>0 S=   Σ_(k≥0) (((−1)^k )/(1+((ak)/b)))(1−ε)^(((ak)/b)+1) Converge uniformily in compact [0,1−ε]  ∫_0 ^1 (dx/(1+x^(a/b) ))=S,(a/b)≠1  since  gcd(a,b)=1   x=tg^(2(b/a)) u  ⇒dx=((2b)/a)(1+tg^2 (u))tg^((2b−a)/a) (u)du  S=((2b)/a).∫_0 ^(π/4) tg^((2b−a)/a) (u)du  S=((2b)/a)∫_0 ^(π/4) sin^(((2b)/a)−1) (u).cos^(((−2b)/a)+2−1) (u)du  withe  B^∼ (x,y,s)=2∫_0 ^s cos^(2x−1) (t).sin^(2y−1) (t)dt  bS=(b/a)B^∼ ((b/a),1−(b/a),(π/4))  ⇒Σ_(k≥0) (((−1)^k )/(ak+b))=((B^∼ ((b/a),1−(b/a),(π/4)))/a)  ,∀(a,b)∈N^(∗2) withe gcd(a,b)=1  and (a,b)≠(1,1)
$$\mathrm{mor}\:\mathrm{generaly} \\ $$$$\mathrm{let}\:\mathrm{S}=\underset{\mathrm{k}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{\mathrm{k}} }{\mathrm{ak}+\mathrm{b}} \\ $$$$\mathrm{withe}\:\:\left(\mathrm{a},\mathrm{b}\right)\in\mathbb{N}^{\ast\mathrm{2}} \\ $$$$\mathrm{withe}\:\mathrm{gcd}\left(\mathrm{a},\mathrm{b}\right)=\mathrm{1} \\ $$$$\mathrm{S}=\frac{\mathrm{1}}{\mathrm{b}}.\Sigma\frac{\left(−\mathrm{1}\right)^{\mathrm{k}} }{\mathrm{1}+\frac{\mathrm{a}}{\mathrm{b}}\mathrm{k}} \\ $$$$\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{1}}{\mathrm{1}+\mathrm{x}^{\frac{\mathrm{a}}{\mathrm{b}}} } \\ $$$$\mid\mathrm{x}\mid<\mathrm{1}\Rightarrow\mathrm{f}\left(\mathrm{x}\right)=\underset{\mathrm{k}\geqslant\mathrm{0}} {\sum}\left(−\mathrm{1}\right)^{\mathrm{k}} .\mathrm{x}^{\frac{\mathrm{ak}}{\mathrm{b}}} \\ $$$$\mathrm{let}\:\varepsilon>\mathrm{0}\Rightarrow\:\:\mathrm{since}\:\mathrm{we}\:\mathrm{have}\:\mathrm{unifirm}\:\mathrm{cv}\:\mathrm{we}\:\mathrm{switch}\:\boldsymbol{\div} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}−\varepsilon} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=\underset{\mathrm{k}\geqslant\mathrm{0}} {\sum}\left(−\mathrm{1}\right)^{\mathrm{k}} \int_{\mathrm{0}} ^{\mathrm{1}−\varepsilon} \mathrm{x}^{\frac{\mathrm{ak}}{\mathrm{b}}} \mathrm{dx} \\ $$$$\mathrm{S}=\underset{\mathrm{k}\geqslant\mathrm{0}} {\sum}\left(−\mathrm{1}\right)^{\mathrm{k}} .\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{ak}}{\mathrm{b}}}\left(\mathrm{1}−\varepsilon\right)^{\frac{\mathrm{ak}}{\mathrm{b}}+\mathrm{1}} \\ $$$$\mathrm{we}\:\mathrm{tack}\:\varepsilon\rightarrow\mathrm{0}\:\:\mathrm{justify}\:\mathrm{by}\:\mathrm{the}\:\mathrm{fact} \\ $$$$\forall\varepsilon>\mathrm{0}\:\mathrm{S}=\:\:\:\underset{\mathrm{k}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{\mathrm{k}} }{\mathrm{1}+\frac{\mathrm{ak}}{\mathrm{b}}}\left(\mathrm{1}−\varepsilon\right)^{\frac{\mathrm{ak}}{\mathrm{b}}+\mathrm{1}} \mathrm{Converge}\:\mathrm{uniformily}\:\mathrm{in}\:\mathrm{compact}\:\left[\mathrm{0},\mathrm{1}−\varepsilon\right] \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{dx}}{\mathrm{1}+\mathrm{x}^{\frac{\mathrm{a}}{\mathrm{b}}} }=\mathrm{S},\frac{\mathrm{a}}{\mathrm{b}}\neq\mathrm{1}\:\:\mathrm{since}\:\:\mathrm{gcd}\left(\mathrm{a},\mathrm{b}\right)=\mathrm{1}\: \\ $$$$\mathrm{x}=\mathrm{tg}^{\mathrm{2}\frac{\mathrm{b}}{\mathrm{a}}} \mathrm{u} \\ $$$$\Rightarrow\mathrm{dx}=\frac{\mathrm{2b}}{\mathrm{a}}\left(\mathrm{1}+\mathrm{tg}^{\mathrm{2}} \left(\mathrm{u}\right)\right)\mathrm{tg}^{\frac{\mathrm{2b}−\mathrm{a}}{\mathrm{a}}} \left(\mathrm{u}\right)\mathrm{du} \\ $$$$\mathrm{S}=\frac{\mathrm{2b}}{\mathrm{a}}.\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{tg}^{\frac{\mathrm{2b}−\mathrm{a}}{\mathrm{a}}} \left(\mathrm{u}\right)\mathrm{du} \\ $$$$\mathrm{S}=\frac{\mathrm{2b}}{\mathrm{a}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{sin}^{\frac{\mathrm{2b}}{\mathrm{a}}−\mathrm{1}} \left(\mathrm{u}\right).\mathrm{cos}^{\frac{−\mathrm{2b}}{\mathrm{a}}+\mathrm{2}−\mathrm{1}} \left(\mathrm{u}\right)\mathrm{du} \\ $$$$\mathrm{withe} \\ $$$$\overset{\sim} {\mathrm{B}}\left(\mathrm{x},\mathrm{y},\mathrm{s}\right)=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{s}} \mathrm{cos}^{\mathrm{2x}−\mathrm{1}} \left(\mathrm{t}\right).\mathrm{sin}^{\mathrm{2y}−\mathrm{1}} \left(\mathrm{t}\right)\mathrm{dt} \\ $$$$\mathrm{bS}=\frac{\mathrm{b}}{\mathrm{a}}\overset{\sim} {\mathrm{B}}\left(\frac{\mathrm{b}}{\mathrm{a}},\mathrm{1}−\frac{\mathrm{b}}{\mathrm{a}},\frac{\pi}{\mathrm{4}}\right) \\ $$$$\Rightarrow\underset{\mathrm{k}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{\mathrm{k}} }{\mathrm{ak}+\mathrm{b}}=\frac{\overset{\sim} {\mathrm{B}}\left(\frac{\mathrm{b}}{\mathrm{a}},\mathrm{1}−\frac{\mathrm{b}}{\mathrm{a}},\frac{\pi}{\mathrm{4}}\right)}{\mathrm{a}}\:\:,\forall\left(\mathrm{a},\mathrm{b}\right)\in\mathbb{N}^{\ast\mathrm{2}} \mathrm{withe}\:\mathrm{gcd}\left(\mathrm{a},\mathrm{b}\right)=\mathrm{1} \\ $$$$\mathrm{and}\:\left(\mathrm{a},\mathrm{b}\right)\neq\left(\mathrm{1},\mathrm{1}\right) \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by mathmax by abdo last updated on 12/Dec/19
pehaps the Q is evaluate Σ_(n=0) ^∞  (((−1)^n )/(5n+1))
$${pehaps}\:{the}\:{Q}\:{is}\:{evaluate}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{5}{n}+\mathrm{1}} \\ $$
Commented by TawaTawa last updated on 12/Dec/19
ok, help me sir
$$\mathrm{ok},\:\mathrm{help}\:\mathrm{me}\:\mathrm{sir} \\ $$
Commented by TawaTawa last updated on 13/Dec/19
God bless you sir
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Commented by mind is power last updated on 13/Dec/19
y′re Welcom
$$\mathrm{y}'\mathrm{re}\:\mathrm{Welcom} \\ $$
Answered by mind is power last updated on 12/Dec/19
diverge +∞
$$\mathrm{diverge}\:+\infty \\ $$
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 +∞
$$\underset{\mathrm{n}\geqslant\mathrm{0}} {\overset{+\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{5n}+\mathrm{1}}\geqslant\underset{\mathrm{n}\geqslant\mathrm{2}} {\sum}\frac{\mathrm{1}}{\mathrm{5n}}=\frac{\mathrm{1}}{\mathrm{5}}\underset{\mathrm{n}\geqslant\mathrm{2}} {\sum}\frac{\mathrm{1}}{\mathrm{n}}……\mathrm{E} \\ $$$$\frac{\mathrm{1}}{\mathrm{n}+\mathrm{1}}\leqslant\int_{\mathrm{n}} ^{\mathrm{n}+\mathrm{1}} \frac{\mathrm{1}}{\mathrm{x}}\leqslant\frac{\mathrm{1}}{\mathrm{n}} \\ $$$$\Rightarrow\underset{\mathrm{n}=\mathrm{2}} {\overset{+\mathrm{N}} {\sum}}\frac{\mathrm{1}}{\mathrm{n}}\geqslant\mathrm{ln}\left(\mathrm{N}+\mathrm{1}\right)−\mathrm{ln}\left(\mathrm{2}\right)\Rightarrow \\ $$$$\mathrm{lim}\underset{\mathrm{N}\rightarrow\infty} {\:}\underset{\mathrm{n}=\mathrm{2}} {\overset{\mathrm{N}} {\sum}}\frac{\mathrm{1}}{\mathrm{n}}\geqslant\underset{\mathrm{N}\rightarrow+\infty} {\mathrm{lim}}\left\{\mathrm{ln}\left(\mathrm{N}+\mathrm{1}\right)−\mathrm{ln}\left(\mathrm{2}\right)\right\}=+\infty \\ $$$$\mathrm{by}\:\mathrm{E}\:\mathrm{we}\:\mathrm{get}\:+\infty \\ $$$$ \\ $$
Commented by TawaTawa last updated on 12/Dec/19
God bless you sir.  Is there shortcut to know a series diverges
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$$$\mathrm{Is}\:\mathrm{there}\:\mathrm{shortcut}\:\mathrm{to}\:\mathrm{know}\:\mathrm{a}\:\mathrm{series}\:\mathrm{diverges}\: \\ $$

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