Question

Use the comparison test to determine whether the following series converge. 194. sum_{n=1}^{infty} a_n where a_n = frac{2}{n(n+1)} 195. sum_{n=1}^{infty} a_n where a_n = frac{1}{n(n+1/2)} 196. sum_{n=1}^{infty} frac{1}{2(n+1)} 197. sum_{n=1}^{infty} frac{1}{2n-1} 198. sum_{n=2}^{infty} frac{1}{(n ln n)^2}

          Use the comparison test to determine whether the following series converge.
194. sum_{n=1}^{infty} a_n where a_n = frac{2}{n(n+1)}
195. sum_{n=1}^{infty} a_n where a_n = frac{1}{n(n+1/2)}
196. sum_{n=1}^{infty} frac{1}{2(n+1)}
197. sum_{n=1}^{infty} frac{1}{2n-1}
198. sum_{n=2}^{infty} frac{1}{(n ln n)^2}

$Use the comparison test to determine whether the following series converge. 194. sumn=1^infty an where an = frac2n(n+1) 195. sumn=1^infty an where an = frac1n(n+1/2) 196. sumn=1^infty frac12(n+1) 197. sumn=1^infty frac12n-1 198. sumn=2^infty frac1(n ln n)^2$

Added by Mark S.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Adi S

06/03/2023

Step 1

For the series $\sum_{n=1}^{\infty} \frac{1}{n(n+1)}$, we can compare it to the series $\sum_{n=1}^{\infty} \frac{1}{n^2}$. Since $\frac{1}{n(n+1)} \leq \frac{1}{n^2}$ for all $n \geq 1$, and we know that the series $\sum_{n=1}^{\infty} \frac{1}{n^2}$ converges Show more…

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