what does (2n^2)/e^(n/5) converge to?

Question

from n=1 to infinity

Answer 1

Assuming you need the sum of the series, I think it converges to about 500.

Initially, the numerator is larger than the denominator but after the 40th term the numerator is smaller than the denominator. This means that the sum of the terms of the series appears to be diverging.

If f(n)=2n²e^-⅕n, f'(n)=(4n-0.4n²)e^-⅕n=0 when the quotient reaches its maximum.

4n=0.4n², 4=0.4n, so at n=10, we have the largest term. After this the terms get progressively smaller and the series is convergent.

This graph is 2n²e^-⅕n, where n is the horizontal axis. The sum of the terms is represented by the area under the curve, although the graph is continuous while the actual sum of the series is based on discrete terms. The graph confirms that n=10 corresponds to the maximum value of the terms.

The integral we need is L=2₁∫^∞n²e^-⅕ndn.

First let's evaluate the general integral K=∫ne^-⅕ndn by integration by parts.

Let u=n, then du=dn; dv=e^-⅕ndn, then v=-5e^-⅕n.

K=uv-∫vdu=-5ne^-⅕n-25e^-⅕n. 

Let J=∫n²e^-⅕ndn and u=n², so du=2ndn, and v and dv are unchanged.

J=-5n²e^-⅕n+10K, so we need to evaluate L=2J=-10n²e^-⅕n+20K between the limits.

That is, L=-10n²e^-⅕n+20(-5ne^-⅕n-25e^-⅕n),

L=(-10n²-100n-500)e^-⅕n.

CHECK:

We need to check this by differentiating:

dL/dn=(-20n-100+2n²+20n+100)e^-⅕n=2n²e^-⅕n. This confirms the general integral.

Applying the limits for L:

L=[(-10n²-100n-500)e^-⅕n]₁^∞=0-(-10-100-500)=-(-610)e^-⅕=610e^-⅕=499.4258 approx.

Note that if the limits had been from n=0, the solution would have been 500.