1. Section 10.5
Integral Test

2. …just don’t forget to eat…
For the next few lessons, we will be concerned with testing many different types of series for convergence.
We will not be particularly concerned with finding the sums of the series that do happen to converge.
If that bothers you,
you may be an engineer.
Feel free to compute the infinite sums of convergent series in your copious spare time.
…just don’t forget to eat…

3. Srīnivāsa Rāmānujan (22 December 1887 – 26 April 1920) was an Indian mathematician and autodidact who, with almost no formal training in pure mathematics, made extraordinary contributions to mathematical analysis, number theory, infinite series and continued fractions. Ramanujan's talent was said by the English mathematician G.H. Hardy to be in the same league as legendary mathematicians such as Gauss, Euler, Cauchy, Newton and Archimedes and he is widely regarded as one of the towering geniuses in mathematics.

4. Born to a poor Brahmin family, Ramanujan first encountered formal mathematics at age 10. He demonstrated a natural ability, and was given books on advanced trigonometry written by S. L. Loney. He mastered them by age 12, and even discovered theorems of his own, including independently re-discovering Euler's identity. He demonstrated unusual mathematical skills at school, winning accolades and awards. By 17, Ramanujan conducted his own mathematical research on Bernoulli numbers and the Euler–Mascheroni constant. He received a scholarship to study at Government College in Kumbakonam, but lost it when he failed his non-mathematical coursework.
Ramanujan joined another college to pursue independent mathematical research, working as a clerk in the Accountant-General's office at the Madras Port Trust Office to support himself. In 1912–1913, he sent samples of his theorems to three academics at the University of Cambridge.

5. Only Hardy recognized the brilliance of his work, subsequently inviting Ramanujan to visit and work with him at Cambridge. He became a Fellow of the Royal Society and a Fellow of Trinity College.
Ramanujan was a devout Hindu and
would not eat meat, nor would he eat food
prepared in any containers that had ever held
any meat. Also, his wife who had stayed
behind in India, told of times when she had to literally put food in his mouth as he would forget to eat for days, entranced by his “sums.” Without his wife’s care and with limited suitable food available in England, Ramanujan died of malnutrition, and possibly liver infection, in 1920 at the age of 32.

6. The integral Test is the first of many tests for convergence.
It is elegant and simple, but unfortunately has limited application.
It only works if the expression can be integrated when 𝑥 replaces 𝑛 in the term that defines the given series.
Example:
1 𝑛 2
→ 𝑥 2 𝑑𝑥
Integral Test is appropriate.
Example:
1 𝑛!
→ 1 𝑥! 𝑑𝑥
Integral Test is NOT appropriate.

7. The terms of the series can form an upper sum.
𝑎 𝑎 𝑎 𝑎 𝑎 𝑎 𝑛
Here’s a graph of a series with 𝑎 𝑛 →0, which therefore MIGHT converge.
The blue curve represents the equivalent expression where 𝑥 replaces 𝑛.
The terms of the series can form an upper sum.
𝑛=1 ∞ 𝑎 𝑛 > 1 ∞ 𝑓(𝑥)𝑑𝑥
If the integral diverges…so does the “larger” series.
The same terms can also form a lower sum.

8. The terms of the series can form an upper sum.
𝑎 𝑎 𝑎 𝑎 𝑎 𝑎 𝑎 𝑛
Here’s a graph of a series with 𝑎 𝑛 →0, which therefore MIGHT converge.
The blue curve represents the equivalent expression where 𝑥 replaces 𝑛.
The terms of the series can form an upper sum.
𝑛=1 ∞ 𝑎 𝑛 > 1 ∞ 𝑓(𝑥)𝑑𝑥
If the integral diverges…so does the “larger” series.
The same terms can also form a lower sum.
𝑛=2 ∞ 𝑎 𝑛 < 1 ∞ 𝑓(𝑥)𝑑𝑥
If the integral converges…so does the series.

11. Because the series can be drawn as either an upper or lower sum, we can conclude that whatever the integral does (converges or diverges), the series does too.

12. Integral Test Given 𝑎 𝑛 is a decreasing positive series and
𝑓 𝑥 a continuous function where 𝑓 𝑛 = 𝑎 𝑛 , then
𝑛=1 ∞ 𝑎 𝑛 and ∞ 𝑓 𝑥 𝑑𝑥
either both converge or both diverge.

13. Ex: #2 (read the directions)
𝑛=1 ∞ 𝑛 𝑒 𝑛 2
Ex: #2 (read the directions)
1 ∞ 𝑥 𝑒 𝑥 2 𝑑𝑥 = ?
∞ 1 𝑒 𝑢 𝑑𝑢
= ∞ 𝑒 −𝑢 𝑑𝑢 ∞ 1
𝑢= 𝑥 2
𝑑𝑢=2𝑥𝑑𝑥
=− 1 2 𝑒 −𝑢
= −1 2∙∞ − −1 2𝑒 = 1 2𝑒
Since the integral converges, so does the series.
(albeit to 𝑎 a value slightly smaller than 1 2𝑒 )

14. Since the integral diverges, so does the series.
𝑛=1 ∞ ln 𝑛 𝑛
Ex: #8 ∞
1 ∞ ln 𝑥 𝑥 𝑑𝑥= ?
0 ∞ 𝑢 𝑑𝑢=
1 2 𝑢 2
=∞−0=∞
𝑢=𝑙𝑛𝑥
𝑑𝑢= 1 𝑥 𝑑𝑥
Since the integral diverges, so does the series.
The only thing that can make these problems difficult
is if the integral itself is difficult.

15. p-Series 1 𝑛 𝑝 Any series of the form where p is a real number.
1 𝑛 𝑝
Any series of the form where p is a real number.
p-series converge if and only if 𝑝>1
Proof
If 𝑝=1, then 𝑛 𝑝 is the divergent harmonic series.
Otherwise, the related integral, 1 ∞ 1 𝑥 𝑝 𝑑𝑥 = 1 −𝑝+1 𝑥 −𝑝+1
If 𝑝>1, then −𝑝+1<0 and 𝑥 −𝑝+1 →0 as 𝑥→∞
In this case, the integral converges to 1 𝑝−1 and the series converges.
If 𝑝<1, then −𝑝+1>0 and 𝑥 −𝑝+1 →∞ as 𝑥→∞
In this case, the integral diverges and therefore the series diverges.
Q.E.D. ∞ 1

16. Ex: #38 (read the directions)
𝑛=3 ∞ 1 𝑛 ln 𝑛 ln⁡( ln 𝑛) 𝑝
Ex: #38 (read the directions) ∞
ln(ln3)
3 ∞ 𝑑𝑥 𝑥 ln 𝑥 ln⁡( ln 𝑥) 𝑝
= ln⁡(𝑙𝑛3) ∞ 𝑑𝑢 𝑢 𝑝
= 1 −𝑝+1 𝑢 −𝑝+1
𝑢=ln⁡( ln 𝑥)
𝑑𝑢= 𝑑𝑥 ⁡𝑥( ln 𝑥)
(we will need a special case if p=1)
= lim 𝑢→∞ 1 (1−𝑝) 𝑢 𝑝−1 − 1 (1−𝑝) ln⁡(𝑙𝑛3) 𝑝−1
This limit will be zero if 𝑝>1, but ∞ if 𝑝<1
if 𝑝=1 then we get 𝑙𝑛⁡(𝑙𝑛3) ∞ 𝑑𝑢 𝑢 = 𝑙𝑛 ∞ − 𝑙𝑛 𝑙𝑛 𝑙𝑛3 =∞
The series converges if and only if 𝑝>1.

17. In the case where all the terms are positive,
If we stop evaluating a convergent positive series after adding n terms, the difference between the true infinite sum and our partial sum is called the error or remainder.
𝑅 𝑛 =𝑆− 𝑆 𝑛
In the case where all the terms are positive,
the error (or remainder) will be positive.
Here’s a representation of the series (Riemann Sum) and it’s related integral for n large.
Drawn as a lower sum, the partial sum stops at the last yellow box.
The error (shown in red) is less than 𝑛 ∞ 𝑓 𝑥 𝑑𝑥 .
Shifted over, drawn as an upper sum, we see that the error is greater than 𝑛+1 ∞ 𝑓 𝑥 𝑑𝑥

18. Integral Test Remainder Estimate
Given 𝑎 𝑛 is a decreasing positive series convergent series
and 𝑓 𝑥 is a continuous function where 𝑓 𝑛 = 𝑎 𝑛
then if 𝑅 𝑛 =𝑆− 𝑆 𝑛
𝑛+1 ∞ 𝑓 𝑥 𝑑𝑥 ≤ 𝑅 𝑛 ≤ 𝑛 ∞ 𝑓 𝑥 𝑑𝑥

19. Ex: #44 (read the directions)
𝑛=1 ∞ 1 𝑛 3 , 𝑘=3
Ex: #44 (read the directions)
We want 𝑅 𝑛 <.0005
It is sufficient is to find 𝑛 such that 𝑛 ∞ 1 𝑥 3 𝑑𝑥 <.0005 ∞ 𝑛
𝑛 ∞ 1 𝑥 3 𝑑𝑥 =− 1 2 𝑥 −2
= 1 ∞ − −1 2 𝑛 2
= 1 2 𝑛 2
1 2 𝑛 2 <.0005
→1<.001 𝑛 2
→1000< 𝑛 2
→𝑛>31.623
The minimum number of terms needed is 32.
𝑆≈𝑆 32 = … ≈
Since this answer is within of 𝑆,
our answer is correct to 3 decimal places.
𝑆≈1.202

20. Enjoy your Homework!