How do you use the integral test to determine the convergence or divergence of #1+1/4+1/9+1/16+1/25+...#?

Our first step is to write the series in series notation. Notice that:

#1+1/4+1/9+1/16+1/25+...#

#=1/1^2+1/2^2+1/3^2+1/4^2+1/5^2+...#

#=sum_(n=1)^oo1/n^2#

So, we want to determine the convergence or divergence of the series #sum_(n=1)^oo1/n^2# using the integral test.

The integral test requires some conditions: for #sum_(n=N)^oof(n)#:

• #f(n)# has to be decreasing on #[N,oo)#
• #f(n)>=0# on the same interval
• #f(n)# is continuous on the interval

For #f(n)# we see that these are all true so the integral test will apply. The actual integral test states that:

The series #sum_(n=N)^oof(n)# converges if and only if the improper integral #int_N^oof(x)dx# converges to a finite number.

So, we want to see if #int_1^oo1/x^2dx# converges to a finite value. When we have an improper integral such as this, replace the infinite bound with a variable, and take the limit as that variable approaches infinity. We write:

#int_1^oo1/x^2dx=lim_(brarroo)int_1^b1/x^2dx#

Then:

#=lim_(brarroo)int_1^bx^-2=lim_(brarroo) [x^-1/(-1)] _ 1^b=lim_(brarroo)[-1/x]_1^b#

Evaluating:

#=lim_(brarroo)(-1/b-(-1/1))=lim_(brarroo)(-1/b+1)#

As #brarroo#, the ratio #-1/brarr0#, so:

#=0+1=1#

So, #int_1^oo1/x^2dx=1#, that is, it converged to a finite value. Thus the series #sum_(n=1)^oo1/n^2=1+1/4+1/9+1/16+1/25+...# converges as well through the integral test.