Pentru a studia convergența acestei serii, mai întâi se va ține cont de faptul că dacă
α
<
0
,
{\displaystyle \alpha <0,\!}
șirul
1
n
α
ln
β
n
{\displaystyle {\frac {1}{n^{\alpha }\ln ^{\beta }n}}\!}
∑
n
≥
2
1
n
α
ln
β
n
,
{\displaystyle \sum _{n\geq 2}{\frac {1}{n^{\alpha }\ln ^{\beta }n}},\!}
este divergentă.
De aceea se presupune că
α
≥
0.
{\displaystyle \alpha \geq 0.\!}
Se vor considera cazurile:
Cazul 1 :
0
<
α
<
1
{\displaystyle 0<\alpha <1\!}
Fie
p
=
α
+
1
2
.
{\displaystyle p={\frac {\alpha +1}{2}}.\!}
α
<
p
<
1.
{\displaystyle \alpha <p<1.\!}
n
p
⋅
1
n
α
ln
β
n
=
n
p
−
α
ln
β
n
.
{\displaystyle n^{p}\cdot {\frac {1}{n^{\alpha }\ln ^{\beta }n}}={\frac {n^{p-\alpha }}{\ln ^{\beta }n}}.\!}
Deoarece
p
−
α
>
0
,
{\displaystyle p-\alpha >0,\!}
lim
n
→
∞
n
p
⋅
1
n
α
ln
β
n
=
∞
.
{\displaystyle \lim _{n\to \infty }n^{p}\cdot {\frac {1}{n^{\alpha }\ln ^{\beta }n}}=\infty .\!}
Astfel, pentru
N
0
≥
2
,
{\displaystyle N_{0}\geq 2,\!}
n
p
⋅
1
n
α
ln
β
n
>
1
,
{\displaystyle n^{p}\cdot {\frac {1}{n^{\alpha }\ln ^{\beta }n}}>1,\!}
ceea ce implică:
1
n
α
ln
β
n
>
1
n
p
.
{\displaystyle {\frac {1}{n^{\alpha }\ln ^{\beta }n}}>{\frac {1}{n^{p}}}.\!}
Deoarece
p
<
1
{\displaystyle p<1\!}
∑
1
n
p
{\displaystyle \sum {\frac {1}{n^{p}}}\!}
∑
n
≥
2
1
n
α
ln
β
n
{\displaystyle \sum _{n\geq 2}{\frac {1}{n^{\alpha }\ln ^{\beta }n}}\!}
este divergentă.
Cazul 2 :
α
>
1
{\displaystyle \alpha >1\!}
Fie
p
=
α
+
1
2
.
{\displaystyle p={\frac {\alpha +1}{2}}.\!}
1
<
p
<
α
.
{\displaystyle 1<p<\alpha .\!}
n
p
⋅
1
n
α
ln
β
n
=
1
n
α
−
p
ln
β
n
.
{\displaystyle n^{p}\cdot {\frac {1}{n^{\alpha }\ln ^{\beta }n}}={\frac {1}{n^{\alpha -p}\ln ^{\beta }n}}.\!}
Deoarece
α
−
p
>
0
,
{\displaystyle \alpha -p>0,\!}
lim
n
→
∞
n
p
⋅
1
n
α
ln
β
n
=
0.
{\displaystyle \lim _{n\to \infty }n^{p}\cdot {\frac {1}{n^{\alpha }\ln ^{\beta }n}}=0.\!}
Astfel, pentru
N
0
≥
2
,
{\displaystyle N_{0}\geq 2,\!}
n
p
⋅
1
n
α
ln
β
n
<
1.
{\displaystyle n^{p}\cdot {\frac {1}{n^{\alpha }\ln ^{\beta }n}}<1.\!}
ceea ce implică:
1
n
α
ln
β
n
<
1
n
p
.
{\displaystyle {\frac {1}{n^{\alpha }\ln ^{\beta }n}}<{\frac {1}{n^{p}}}.\!}
Seria
∑
1
n
p
{\displaystyle \sum {\frac {1}{n^{p}}}\!}
p
>
1.
{\displaystyle p>1.\!}
∑
n
≥
2
1
n
α
ln
β
n
{\displaystyle \sum _{n\geq 2}{\frac {1}{n^{\alpha }\ln ^{\beta }n}}\!}
este convergentă.
Cazul 3 :
α
=
1
{\displaystyle \alpha =1\!}
Considerăm funcția:
f
(
x
)
=
1
x
ln
β
x
.
{\displaystyle f(x)={\frac {1}{x\ln ^{\beta }x}}.\!}
E ușor de verificat că, pentru un x suficient de mare (mai exact
x
>
e
−
β
{\displaystyle x>e^{-\beta }\!}
f
(
x
)
{\displaystyle f(x)\!}
∫
M
n
+
1
f
(
x
)
d
x
≤
1
(
M
+
1
)
ln
β
(
M
+
1
)
+
⋯
+
1
n
ln
β
n
,
{\displaystyle \int _{M}^{n+1}f(x)dx\leq {\frac {1}{(M+1)\ln ^{\beta }(M+1)}}+\cdots +{\frac {1}{n\ln ^{\beta }n}},\!}
și
1
M
ln
β
M
+
⋯
+
1
(
n
−
1
)
ln
β
(
n
−
1
)
≤
∫
M
n
−
1
f
(
x
)
d
x
,
{\displaystyle {\frac {1}{M\ln ^{\beta }M}}+\cdots +{\frac {1}{(n-1)\ln ^{\beta }(n-1)}}\leq \int _{M}^{n-1}f(x)dx,\!}
unde M este un număr întreg astfel ales încât f(x) este descrescătoare pe
(
M
,
∞
)
.
{\displaystyle (M,\infty ).\!}
β
≠
1
,
{\displaystyle \beta \neq 1,\!}
∫
M
n
+
1
1
x
ln
β
x
d
x
=
[
1
1
−
β
ln
1
−
β
x
]
M
n
+
1
=
ln
1
−
β
(
n
+
1
)
−
ln
1
−
β
M
1
−
β
,
{\displaystyle \int _{M}^{n+1}{\frac {1}{x\ln ^{\beta }x}}dx={\bigg [}{\frac {1}{1-\beta }}\ln ^{1-\beta }x{\bigg ]}_{M}^{n+1}={\frac {\ln ^{1-\beta }(n+1)-\ln ^{1-\beta }M}{1-\beta }},\!}
și dacă
β
=
1
{\displaystyle \beta =1\!}
M
=
2
{\displaystyle M=2\!}
∫
2
n
+
1
1
x
ln
x
d
x
=
[
ln
(
ln
x
)
]
2
n
+
1
=
ln
(
ln
(
n
+
1
)
)
−
ln
(
ln
2
)
.
{\displaystyle \int _{2}^{n+1}{\frac {1}{x\ln x}}dx={\bigg [}\ln(\ln x){\bigg ]}_{2}^{n+1}=\ln(\ln(n+1))-\ln(\ln 2).\!}
Cazul a. :
β
<
1
,
{\displaystyle \beta <1,\!}
ln
1
−
β
(
n
+
1
)
−
ln
1
−
β
M
1
−
β
≤
1
(
M
+
1
)
ln
β
(
M
+
1
)
+
⋯
+
1
n
ln
β
n
.
{\displaystyle {\frac {\ln ^{1-\beta }(n+1)-\ln ^{1-\beta }M}{1-\beta }}\leq {\frac {1}{(M+1)\ln ^{\beta }(M+1)}}+\cdots +{\frac {1}{n\ln ^{\beta }n}}.\!}
Deoarece
lim
n
→
∞
ln
1
−
β
(
n
+
1
)
−
ln
1
−
β
M
1
−
β
=
∞
,
{\displaystyle \lim _{n\to \infty }{\frac {\ln ^{1-\beta }(n+1)-\ln ^{1-\beta }M}{1-\beta }}=\infty ,\!}
∑
1
n
ln
β
n
{\displaystyle \sum {\frac {1}{n\ln ^{\beta }n}}\!}
Cazul b. :
β
>
1
,
{\displaystyle \beta >1,\!}
1
M
ln
β
M
+
⋯
+
1
(
n
−
1
)
ln
β
(
n
−
1
)
≤
ln
1
−
β
(
n
+
1
)
−
ln
1
−
β
M
1
−
β
,
{\displaystyle {\frac {1}{M\ln ^{\beta }M}}+\cdots +{\frac {1}{(n-1)\ln ^{\beta }(n-1)}}\leq {\frac {\ln ^{1-\beta }(n+1)-\ln ^{1-\beta }M}{1-\beta }},\!}
dar, deoarece:
ln
1
−
β
(
n
+
1
)
−
ln
1
−
β
M
1
−
β
<
1
β
−
1
,
{\displaystyle {\frac {\ln ^{1-\beta }(n+1)-\ln ^{1-\beta }M}{1-\beta }}<{\frac {1}{\beta -1}},\!}
pentru valori mari ale lui n , obținem:
1
M
ln
β
M
+
⋯
+
1
(
n
−
1
)
ln
β
(
n
−
1
)
<
1
β
−
1
,
{\displaystyle {\frac {1}{M\ln ^{\beta }M}}+\cdots +{\frac {1}{(n-1)\ln ^{\beta }(n-1)}}<{\frac {1}{\beta -1}},\!}
ceea ce înseamnă că șirul sumelor parțiale asociate seriei:
∑
n
≥
2
1
n
ln
β
n
{\displaystyle \sum _{n\geq 2}{\frac {1}{n\ln ^{\beta }n}}\!}
este marginit.
Deci seria este convergentă.
Cazul c. :
β
=
1
,
{\displaystyle \beta =1,\!}
∫
2
n
+
1
x
ln
x
d
x
≤
1
3
ln
3
+
⋯
+
1
n
ln
n
,
{\displaystyle \int _{2}^{n+1}{\frac {x\ln x}{d}}x\leq {\frac {1}{3\ln 3}}+\cdots +{\frac {1}{n\ln n}},\!}
ceea ce implică:
ln
(
ln
(
n
+
1
)
)
−
ln
(
ln
2
)
≤
1
3
ln
3
+
⋯
+
1
n
ln
n
.
{\displaystyle \ln(\ln(n+1))-\ln(\ln 2)\leq {\frac {1}{3\ln 3}}+\cdots +{\frac {1}{n\ln n}}.\!}
Dar cum:
lim
n
→
∞
ln
(
ln
(
n
+
1
)
)
−
ln
(
ln
2
)
=
∞
,
{\displaystyle \lim _{n\to \infty }\ln(\ln(n+1))-\ln(\ln 2)=\infty ,\!}
ajungem la concluzia că șirul sumelor parțiale asociat seriei:
∑
n
≥
2
1
n
ln
n
{\displaystyle \sum _{n\geq 2}{\frac {1}{n\ln n}}\!}
nu este mărginit.
Deci seria nu este divergentă.
![{\displaystyle \int _{M}^{n+1}{\frac {1}{x\ln ^{\beta }x}}dx={\bigg [}{\frac {1}{1-\beta }}\ln ^{1-\beta }x{\bigg ]}_{M}^{n+1}={\frac {\ln ^{1-\beta }(n+1)-\ln ^{1-\beta }M}{1-\beta }},\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/03b4cdb6538c28ba422ca2229bb378b6e9c5f818)
![{\displaystyle \int _{2}^{n+1}{\frac {1}{x\ln x}}dx={\bigg [}\ln(\ln x){\bigg ]}_{2}^{n+1}=\ln(\ln(n+1))-\ln(\ln 2).\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c6ef7fbfeaf9f7abc2517eb7d6bec07f0fd4dfea)
