Determine whether the series is convergent or divergent.
1. sum (n=1, infinity) n/2n^3+4n+2
2. sum (n=1, infinity) ne^(-n^2)
3. sum (n=1, infinity) 2^n/(2n)!
4. sum (n=1, infinity) 3/sqrt(n^3+2n+1)
Please state the test used.
Please Help!!!! Thanks so much.
1. sum (n=1, infinity) n/2n^3+4n+2
2. sum (n=1, infinity) ne^(-n^2)
3. sum (n=1, infinity) 2^n/(2n)!
4. sum (n=1, infinity) 3/sqrt(n^3+2n+1)
Please state the test used.
Please Help!!!! Thanks so much.
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1) Using the Comparison Test:
Since n/(2n^3 + 4n + 2) < n/(1n^3 + 0) = 1/n^2, and Σ(n=1 to ∞) 1/n^2 is a convergent p-series,
we conclude that Σ(n=1 to ∞) n/(2n^3 + 4n + 2) also converges.
2) Using the Integral Test:
Since ∫(x = 1 to ∞) xe^(-x^2) dx = (-1/2)e^(-x^2) {for x = 1 to ∞} = 1/2 (convergent),
we conclude that Σ(n=1 to ∞) ne^(-n^2) also converges.
3) Using the Ratio Test:
r = lim(n→∞) [2^(n+1)/(2n+2)!] / [2^n/(2n)!]
..= lim(n→∞) 2/[(2n+2)(2n+1)]
..= 0.
Since r = 0 < 1, the series converges.
4) Using the Comparison Test:
Since 3/√(n^3+2n+1) < 3/√(n^3) = 3/n^(3/2), and Σ(n=1 to ∞) 3/n^(3/2) is convergent (multiple of a convergent) p-series, we conclude that Σ(n=1 to ∞) 3/√(n^3+2n+1) also converges.
I hope this helps!
Since n/(2n^3 + 4n + 2) < n/(1n^3 + 0) = 1/n^2, and Σ(n=1 to ∞) 1/n^2 is a convergent p-series,
we conclude that Σ(n=1 to ∞) n/(2n^3 + 4n + 2) also converges.
2) Using the Integral Test:
Since ∫(x = 1 to ∞) xe^(-x^2) dx = (-1/2)e^(-x^2) {for x = 1 to ∞} = 1/2 (convergent),
we conclude that Σ(n=1 to ∞) ne^(-n^2) also converges.
3) Using the Ratio Test:
r = lim(n→∞) [2^(n+1)/(2n+2)!] / [2^n/(2n)!]
..= lim(n→∞) 2/[(2n+2)(2n+1)]
..= 0.
Since r = 0 < 1, the series converges.
4) Using the Comparison Test:
Since 3/√(n^3+2n+1) < 3/√(n^3) = 3/n^(3/2), and Σ(n=1 to ∞) 3/n^(3/2) is convergent (multiple of a convergent) p-series, we conclude that Σ(n=1 to ∞) 3/√(n^3+2n+1) also converges.
I hope this helps!