Integral Approximation to Infinite sum Vs Cauchy's first theorem

Question

$$ \lim_{n\to \infty}\sum_{i=0}^n 1/(3*n +i) $$ . After applying cauchy's first theorem on pints, I get the answer as 1/4 , but after expressing the above sum as a definite integral I get the answer as log(4/3). Why do I get two different answers?

Answer 1

"Cauchy's First Theorem on limits" seems to be the name (in perhaps indian curriculum based on my google searches?) given to the regularity of Cesaro summation. I'll quote it here since you didn't state it.

(Cauchy's First Theorem on limits) That is, if $a_n \to a$, then also $\frac{1}{n+1} \sum_{i=0}^n a_i = \frac{a_1 + \dots + a_n }{n+1} \to a$ as well.

Your partial sums are $$ \sum_{i=0}^n a_i = \frac1{n+1} \sum_{i=0}^n \frac{n+1}{3n+i} $$

The candidate to directly apply the theorem would be $a_i = \frac{n+1}{3n+i}$, but this is not a function only of $i$; it depends on $n$, so the theorem does not apply. What you can say though, is that

$$ \frac{n+1}{3n+i} ≥ \frac{n}{3n+n} = \frac{1}{4}$$ and this right hand side was your supposed $a_n$. The sum of this lower bound can be computed with or without Cauchy's first theorem and so (if the limit exists) $$ \lim_{n\to \infty} \frac1{n+1} \sum_{i=0}^n \frac{n+1}{3n+i} \ge \frac14.$$ Note that $\frac14 = 0.25 \le \log(4/3) \approx 0.287$ which is consistent (assuming that your Riemann sum work is correct and that by $\log$ you mean the natural logarithm.)

In the post I linked, he attempted your solution method in his question 2, which is $$ \frac{1}n \sum_{i=1}^n \frac{1}{n(1+\frac in)^2} $$ so his candidate is $a_i = \frac{1}{n(1+\frac in)^2}$. Here, $(1+\frac in)^2 ≤ 2^2 = 4$, so by right, he should only get the lower bound (should the limit exist) $$ \lim_{n\to\infty} \frac{1}n \sum_{i=1}^n \frac{1}{n(1+\frac in)^2} \ge 0.$$ He could have instead used the simple bound $$ (1+\frac in)^2 \ge 1$$ to get $$ \frac{1}n \sum_{i=1}^n \frac{1}{n(1+\frac in)^2} \le \frac{1}{n} \sum_{i=1}^n \frac{1}n = \frac{1}n \to 0.$$ In his case the answer turns out to be right but in your case the answer isn't, and this is because the theorem above should not have been applied.

PS see also Confusion on Cauchy's first theorem on limits