Question

Determine the convergence or divergence of each series. a. \(\sum_{n=1}^{\infty} \frac{n+1}{3n+1}\) b. \(\sum_{n=1}^{\infty} \left(\frac{\pi}{6}\right)^n\) c. \(\sum_{n=1}^{\infty} ne^{-n^2}\) d. \(\sum_{n=1}^{\infty} \frac{1}{3n+1}\) e. \(\sum_{n=1}^{\infty} (-1)^n \frac{3}{4n+1}\) f. \(\sum_{n=1}^{\infty} \frac{n!}{10^n}\) g. \(\sum_{n=1}^{\infty} \left(\frac{n+1}{2n+1}\right)^n\)

          Determine the convergence or divergence of each series.
a. \(\sum_{n=1}^{\infty} \frac{n+1}{3n+1}\)
b. \(\sum_{n=1}^{\infty} \left(\frac{\pi}{6}\right)^n\)
c. \(\sum_{n=1}^{\infty} ne^{-n^2}\)
d. \(\sum_{n=1}^{\infty} \frac{1}{3n+1}\)
e. \(\sum_{n=1}^{\infty} (-1)^n \frac{3}{4n+1}\)
f. \(\sum_{n=1}^{\infty} \frac{n!}{10^n}\)
g. \(\sum_{n=1}^{\infty} \left(\frac{n+1}{2n+1}\right)^n\)

Determine the convergence or divergence of each series.
a. ∑n=1^∞(n+1)/(3n+1)
b. ∑n=1^∞((π)/(6))^n
c. ∑n=1^∞ ne^-n^2
d. ∑n=1^∞(1)/(3n+1)
e. ∑n=1^∞ (-1)^n (3)/(4n+1)
f. ∑n=1^∞(n!)/(10^n)
g. ∑n=1^∞((n+1)/(2n+1))^n

Added by Zachary O.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Israel Hernandez

Step 1

n+1 / (3n+1) To determine the convergence or divergence of this series, we can use the limit comparison test. Let's compare it to the series 1/n. lim (n→∞) (n+1 / (3n+1)) / (1/n) = lim (n→∞) (n+1) / (3n+1) * (n/1) = lim (n→∞) (n+1) * (n/1) / (3n+1) = lim (n→∞) Show more…

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Determine the convergence or divergence of each series: a. n+1 / (3n+1) b. e-n^2 c. 1 / (d * M8 * 3n+1) d. 3 / (n! * 10") e. 4n+1 f. iM8 / (n+1) g. 2n+1