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Determine whether or not the series converges or diverges. Any test is accepted.

1. &sum (infinity on top; n=1 on bottom) times (3^(2n)/n!)

2. ∑ (infinity on top; n=1 on bottom) times (1/(2+5^n))

3. ∑ (infinity on top; n=1 on bottom) times (3n^3/2n+1)^n

4. ∑ (infinity on top; n=1 on bottom) times (1/(2 sqrt(n)).

2007-07-16 19:09:06 · 1 answers · asked by Kayla G 1 in Science & Mathematics Mathematics

1 answers

1. Diverges. Each new term is 9/n. This is greater than 1/n, which diverges.
2. Converges. Can be shown to have a limiting value.
3. Think this is messed up.
4. Diverges. Terms are greater than 1/n, which diverges.

2007-07-16 19:26:33 · answer #1 · answered by cattbarf 7 · 0 0

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