How do you use the comparison test for #sum 1/(5n^2+5)# for n=1 to #n=oo#?

Question

Charles Arocha · Accepted Answer

The Direct Comparison Test (not to be confused with the Limit Comparison Test) says, paraphrased: Choose some series #suma_n# and some other series #sumb_n# such that #a_n >= 0# and #b_n >= 0#, while #a_n <= b_n#. When that is done, the following limit is said to be a valid test: If #sumb_n# is convergent, then #suma_n# is as well.
If #suma_n# is divergent, then #sumb_n# is as well. Notice how the larger series determines convergence, not the smaller one. It's exactly like this: If b, then a.
If not a, then not b. which is a statement and then its contrapositive, which must be true. The affirmative is the convergence, and the denial is the divergence. This is not true: If b, then a.
If a, then b. (If #a_n# converges, that says nothing about #b_n#.) Now that that's out of the way, let's pick #a_n# and #b_n#. Picking #a_n# is easy; it's your series. #b_n#, however, must be always greater than #a_n#, and also have a lower bound at #0# just like #b_n#.  An easy way to pick #b_n# is to let #b_n# be #a_n# with the far-right constant smaller, since the denominator will now be smaller and the overall would be larger. #suma_n = sum_(n=1)^(oo)  1/(5n^2 + 5)#
#sumb_n = sum_(n=1)^(oo)  1/(5n^2 + 4)# Notice that since we're going towards #oo#, the constants don't even matter by then.  Now we need to show whether #b_n# is convergent, or whether #a_n# is divergent. As a form of a p-series (#sum 1/(n^p + k)# converges when #p > 1#), there's no doubt it's going to converge, but let's just perform the sum to see how it goes.  #1/(5(1)^2 + 4) + 1/(5(2)^2 + 4) + 1/(5(3)^2 + 4) + 1/(5(4)^2 + 4)#
# = 1/9 + 1/24 + 1/49 + 1/84 + ...# #= 0.bar11 + 0.041bar6 + 0.0204 + 0.0119 + ...# The denominator is approaching 0 pretty quickly in the first few values of #n#, and will continue rather quickly at each successive #n# (#n^2# gets very steep pretty quickly). So therefore, #a_n# must converge as well. Wolfram Alpha says the convergence result is #pi/10cothpi - pi/10#, so there is indeed a finite sum for #a_n#. The sum for #b_n# is similar in magnitude but you can type that into Wolfram Alpha and see the exact answer if you're curious.

Charles Anctil · Answer

undefined To use the comparison test for the series $\sum \frac{1}{5n^2 + 5}$ from $n = 1$ to $n = \infty$, we need to find a series whose terms are less than or equal to the terms of the given series and whose sum is known.

We can rewrite the given series as $\sum \frac{1}{5(n^2 + 1)}$. Now, since $n^2 + 1 > n^2$ for all $n \geq 1$, we have $\frac{1}{5(n^2 + 1)} < \frac{1}{5n^2}$ for all $n \geq 1$.

The series $\sum \frac{1}{5n^2}$ is a well-known convergent series, known as a p-series with $p = 2$. Its sum is $\frac{1}{5} \times \frac{\pi^2}{6}$.

By the comparison test, since $\frac{1}{5(n^2 + 1)} < \frac{1}{5n^2}$ for all $n \geq 1$ and $\sum \frac{1}{5n^2}$ converges, then $\sum \frac{1}{5(n^2 + 1)}$ also converges.