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

Question

Christopher Agresta · Accepted Answer

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)#:

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.

Isaiah Allen · Answer

undefined To use the integral test to determine the convergence or divergence of the series 1 + 1/4 + 1/9 + 1/16 + 1/25 + ..., we need to consider the corresponding function $ f(x) = \frac{1}{x^2} $.

1. First, note that $ f(x) $ is positive, continuous, and decreasing for $ x \geq 1 $.
2. Then, we integrate $ f(x) $ from 1 to infinity. This gives us $ \int_{1}^{\infty} \frac{1}{x^2} \, dx = 1 $.
3. Since the integral converges, by the integral test, the series $ \sum_{n=1}^{\infty} \frac{1}{n^2} $ also converges.

Therefore, the series 1 + 1/4 + 1/9 + 1/16 + 1/25 + ... converges.