I wish to prove divergence of
∞∑n=3√n+2n−2
I wish to do so by comparison, since n≥3:
∞∑n=3√n+2n−2>∞∑n=31+2n−2>∞∑n=33n>∞∑n=31n→∞
And the harmonic series is divergent, so if we just remove finitely many terms, we still have that it is divergent, because divergence is determined "in the tail". We have a divergent minorant series and hence the original series diverges to ∞.
Is this approach fine, or is there some more elegant method, this was about the simplest thing I could think of.
Alternatively we have:
∞∑n=3√n+2n−2>∞∑n=3√n+2n=∞∑n=31√n+2n→∞
Answer
Not clear what your question is, but your answer is correct.
Your approach is fine. Comparison test would be the proper test to use.
One can also show that ∞∑n=3√n+2n−2≥∞∑n=3√n+2n
and show that the rightmost sum is diverging via the integral test.
No comments:
Post a Comment