Use the integral test to determine whether the series converges.

Question

Σ^∞_n=1  1/5ⁿ

Answer 1

We compare this with the integral ∫^∞₁(1/5^x)dx = ∫^∞₁(5^-x)dx.  

 

We know that in general, ∫a^xdx = a^x /ln(a) + C.  To solve ∫a^-xdx we can make a u-substitution:

 

u=-x

du = -dx

-du = dx, and the integral then becomes

-∫a^udu, which is -a^u/ln(a) + C.  Since u = -x, our the solution to the integral is

 

∫a^-xdx = -a^-x/ ln(a) + C.  

 

In this problem, a = 5, and since it's a definite integral, the constant of integration subtracts out to zero and disappears.

 

We need to make one adjustment; integration with a limit at infinity is an improper integral.  Instead, we integrate with N as an upper limit of integration, and take the limit of the integral as N→∞.

 

Putting all of this together, we have

lim_N→∞ ∫^N₁(5^-x)dx

= lim_N→∞ [-5^-x /ln(5)]^N₁; reverse the sign of the expression, and interchange upper and lower limits:

=lim_N→∞[5^-x/ln(5)]¹_N

=5^-1/ln(5) - lim_N→∞5^-Nln(5)

=1/5ln(5) - 0

=1/5ln(5).

 

Since the integral converges, so does the series.