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$$ \lim_{n\to \infty}\sum_{i=0}^n 1/(3*n +i) $$ . 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?

Jor_El
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  • Cauchy's first theorem on limits – Jor_El Feb 01 '19 at 13:27
  • I suppose this is your Cauchy's first theorem, https://math.stackexchange.com/questions/1930373/limits-and-cauchys-first-theorem

    Can you show your work for applying this theorem?

     – Calvin Khor Feb 01 '19 at 13:27
  • Yes the same. I multiplied and divided by 'n' after which a_n became (n/4n). – Jor_El Feb 01 '19 at 13:31
  • Is it just a direct application of the theorem? Because your partial sum is $$ \frac1n\sum_{i=0}^n \frac n{3n+i} $$ which is not $$ \frac1n \sum_{i=0}^n a_i = \frac1n \sum_{i=0}^n \frac i{4i} = \frac1n \sum_{i=0}^n \frac 1{4}$$ – Calvin Khor Feb 01 '19 at 13:52
  • By a_n I meant the n_th term in the partial sum. – Jor_El Feb 01 '19 at 13:55
  • Oh, you mean the last term in the sum? Is that how it works...? – Calvin Khor Feb 01 '19 at 13:58
  • Yes. It's been done the same way in the link you have provided. – Jor_El Feb 01 '19 at 14:03
  • I see that, but I don't believe it. If it is true, it doesn't follow immediately from the version stated at the top, which is $$ a_n \to a \implies \frac1n\sum_{i=0}^n a_i \to a $$ – Calvin Khor Feb 01 '19 at 14:04
  • Is there any exception to the theorem? How do I choose between the two methods? – Jor_El Feb 01 '19 at 14:06

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"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

Calvin Khor
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