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Practice Question
Related Topics
Wize University Calculus 2 Textbook > Sequences and Series
Root Test
3 Activities
Practice Question
Determine whether
∑
n
=
1
∞
3
2
n
−
1
n
n
\sum\limits^\infty_{n=1}\frac{3^{2n-1}}{n^n}
n
=
1
∑
∞
n
n
3
2
n
−
1
converges or diverges.
It converges absolutely
It converges conditionally
It diverges
Cannot be determined using our known tests
I don't know
Check Submission
More Root Test Questions:
Practice: Root Test
Detemine whether the series
∑
n
=
2
∞
(
−
1
)
n
(
ln
n
)
n
{\displaystyle \sum_{n=2}^\infty}\frac{(-1)^n}{(\ln n)^n}
n
=
2
∑
∞
(
l
n
n
)
n
(
−
1
)
n
converges absolutely, converges conditionally, or diverges.
Practice: Root Test
Detemine whether the series
∑
n
=
2
∞
(
−
1
)
n
(
ln
n
)
n
{\displaystyle \sum_{n=2}^\infty}\frac{(-1)^n}{(\ln n)^n}
n
=
2
∑
∞
(
l
n
n
)
n
(
−
1
)
n
converges absolutely, converges conditionally, or diverges.
Practice: Root Test
Practice: Root Test
Detemine whether the series
∑
n
=
2
∞
(
−
1
)
n
(
ln
n
)
n
{\displaystyle \sum_{n=2}^\infty}\frac{(-1)^n}{(\ln n)^n}
n
=
2
∑
∞
(
l
n
n
)
n
(
−
1
)
n
converges absolutely, converges conditionally, or diverges.
Practice: Root Test
Practice: Root Test
Detemine whether the series
∑
n
=
2
∞
(
−
1
)
n
(
ln
n
)
n
{\displaystyle \sum_{n=2}^\infty}\frac{(-1)^n}{(\ln n)^n}
n
=
2
∑
∞
(
l
n
n
)
n
(
−
1
)
n
converges absolutely, converges conditionally, or diverges.
Practice: Root Test
Practice: Root Test
Detemine whether the series
∑
n
=
2
∞
(
−
1
)
n
(
ln
n
)
n
{\displaystyle \sum_{n=2}^\infty}\frac{(-1)^n}{(\ln n)^n}
n
=
2
∑
∞
(
l
n
n
)
n
(
−
1
)
n
converges absolutely, converges conditionally, or diverges.
Practice Question
Practice Question
Determine whether
∑
n
=
1
∞
3
2
n
−
1
n
n
\sum\limits^\infty_{n=1}\frac{3^{2n-1}}{n^n}
n
=
1
∑
∞
n
n
3
2
n
−
1
converges or diverges.
Practice: Root Test
Practice: Root Test
Detemine whether the series
∑
n
=
2
∞
(
−
1
)
n
(
ln
n
)
n
{\displaystyle \sum_{n=2}^\infty}\frac{(-1)^n}{(\ln n)^n}
n
=
2
∑
∞
(
l
n
n
)
n
(
−
1
)
n
converges absolutely, converges conditionally, or diverges.