determine whether{ncosn/n^2+1) converges or diverges.If it vonverges, find the limit.?

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answers:
Jeffrey K say: As n gets large, the denominator gets much bigger than the numerator so a small number over a bigger number approaches 0.
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kb say: Use the Squeeze Theorem, noting that -1 ≤ cos n ≤ 1 for all n.

Thus, we have for all n:
-n/(n²+1) ≤ n cos(n)/(n²+1) ≤ n/(n²+1).

Since lim(n→∞) ±n/(n²+1) = 0, we conclude by the Squeeze
Theorem that lim(n→∞) n cos(n)/(n²+1) = 0.

(In other words, the sequence {n cos(n)/(n²+1)} converges to 0.)

I hope this helps!
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Ian H say: Consider y = [x/(x^2+1)]cos(x). This graph suggests convergence

https://www.wolframalpha.com/input/?i=y+...

y = x/(x ^2 + 1) becomes like y = 1/x for large x.
The cosine of x oscillates about those values giving smaller alternating values for y as x tends to infinity.
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