Consider the series $$\lim \limits_{n \to \infty}\sum_{i=1}^n \frac{n}{n^2+i}$$

Considering individual terms, we get $$\lim \limits_{n \to \infty}\frac{n}{n^2+i}=\lim \limits_{n \to \infty}\frac{1}{n+\frac{i}{n}}=0.$$ Hence the sum would be $0$.

However, considering an alternative approach, we observe that $\frac{n}{n^2+n}<\frac{n}{n^2+i}<\frac{n}{n^2+1}$. Therefore $$\sum_{i=1}^n \frac{n}{n^2+n}<\sum_{i=1}^n \frac{n}{n^2+i}<\sum_{i=1}^n \frac{n}{n^2+1} \implies \frac{n^2}{n^2+n}<\sum_{i=1}^n \frac{n}{n^2+i}<\frac{n^2}{n^2+1}$$ Taking limit $\lim \limits_{n \to \infty}$ on the three expressions, we find that $\lim \limits_{n \to \infty}\frac{n^2}{n^2+n}=\lim \limits_{n \to \infty}\frac{n^2}{n^2+1}=1.$

Therefore by the sandwich theorem, $\lim \limits_{n \to \infty}\sum_{i=1}^n \frac{n}{n^2+i}=1$

Since there exists a unique limit, there must be only one solution. Which approach is the correct one?

There was one suggested reason that each of the individual limits that was calculated in the 1st approach had to be summed up to infinite terms. Since each of the limits were infintesimally small, we cannot predict the way they will behave when summed to infinity and therefore the approach is incorrect. However, my counter was that the limits are exact and are not infinitesimals. Therefore 0 can be summed up infinitely many times and the answer will still be zero.

Is this reasoning correct/wrong? What must be the answer to the problem?


2 Answers 2


The second solution you have is correct (although in the last step, the expression on the right side of the inequality should just be $n^2/(n^2 + 1)$). The first method is incorrect; since the number of terms of the sum varies with $n$, you cannot simply use the addition rule for limits.

  • $\begingroup$ Yes, sum of infinite number of infinite tiny terms is not necessarily zero. $\endgroup$
    – Arashium
    Feb 13, 2015 at 17:16
  • $\begingroup$ It's not nearly zero, it's exact 0 $\endgroup$
    – user117913
    Feb 13, 2015 at 17:32
  • 2
    $\begingroup$ That's just it though @user117913, although the limit of the terms is exactly zero, the terms themselves aren't exactly zero but are instead nearly zero. Two common examples are $\lim\limits_{n\to\infty} \sum\limits_{i=1}^n \frac{1}{2^i} = 1$ and $\lim\limits_{n\to\infty} \sum\limits_{i=1}^n \frac{1}{i} = \infty$. It is possible that the sum of infinitely many nearly zero entries is finite, infinite, zero, and anywhere inbetween. In your case in particular, each term is positive and nonzero, which means that the sum is positive and nonzero. $\endgroup$
    – JMoravitz
    Feb 13, 2015 at 17:35

Just an addition: you can get some nice asymptotics out of this sum (set $t=n^2$): $$ n \sum_{k=0}^{n} \frac{1}{t+k} = n \bigg(\frac{1}{t} + \ldots + \frac{1}{t+n} \bigg) \\ =n \bigg(1 + \frac{1}{2} + \ldots \frac{1}{t+n} -(1+\frac{1}{2} +\ldots \frac{1}{t-1} \bigg)\\ \sim n (\log (t+n) - \log t) = n \log (1+\frac{1}{n}) $$ which, for large $n$ converges to 1.


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