Find the limits of the following sequences ∑n=1∞n⋅ln⁡(n)n2+1 ∑n=1∞sin⁡(n)n

We have, by applying the Squeeze Theorem limn→∞sin⁡(n)n=0 We have limn→∞n⋅ln⁡(n)n2+1 =limx→∞ln⁡(n)n1+1n2 This implies =limn→∞ln⁡(n)nlimn→∞1+1n2 Since limn→∞ln⁡(n)n=0 =01=0

"Find the limits of the following sequences ∑n=1∞n⋅ln⁡(n)n2+1..." was answered by our experts

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sanuluy

Answered question

2021-08-22

Find the limits of the following sequences
$\sum_{n = 1}^{\infty} \frac{n \cdot \ln (n)}{n^{2} + 1}$
$\sum_{n = 1}^{\infty} \frac{\sin (n)}{n}$

Answer & Explanation

Theodore Schwartz

Skilled2021-08-23Added 99 answers

We have, by applying the Squeeze Theorem

lim_{n \to \infty} \frac{\sin (n)}{n} = 0

We have

lim_{n \to \infty} \frac{n \cdot \ln (n)}{n^{2} + 1}

= lim_{x \to \infty} \frac{\frac{\ln (n)}{n}}{1 + \frac{1}{n^{2}}}

This implies

= \frac{lim_{n \to \infty} \frac{\ln (n)}{n}}{lim_{n \to \infty} 1 + \frac{1}{n^{2}}}

Since

lim_{n \to \infty} \frac{\ln (n)}{n} = 0

= \frac{0}{1} = 0

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