Test each following to convergance
∑(n=1,∞)(√(n^3 +5n -1)/(n^2 - sin(n^3))

2. the sequence { (n^2 /(3*n +1 ))sin ((\pi[/tex])/(3*n ) } for all n≥1

pi =3.14.

1. Compare the series to the divergent p-series,

Use the limit comparison test: we have

so the given series also diverges.

Breakdown for the limit: oscillates between -1 and 1, so its contribution to the denominator is neglible and we can replace it with . In the square root in the numerator, the term dominates, so . So the given summand is approximated by

2. Let , so that as . Then

Recall that as . Rewrite the limit as

Then , and we're left with

which is continuous at , giving a limit of .

i am currently in the exact same situation rn. algebra2/adv algebra is killing

step-by-step explanation:

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