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On trying to compute the following limit, $$\lim_{n\to \infty}\bigg({\frac{1}{n^2+n+1}+\frac{2}{n^2+n+2}+\dots+\frac{n}{n^2+n+n}\bigg)}$$ I used the well-known addition property of limits to conclude that the above expression is equal to $$\lim_{n\to \infty}{\bigg(\frac{1}{n^2+n+1}\bigg)}+\lim_{n\to \infty}{\bigg(\frac{2}{n^2+n+2}\bigg)}+\dots+\lim_{n\to \infty}{\bigg(\frac{n}{n^2+n+n}\bigg)}$$ Now, since reach of the individual limits tends to $0$ as $n$ tends to infinity, I concluded that the final answer is $0$.

But the solution given to this problem uses the Squeeze theorem and states that the limit is $\frac{1}{2}$, and so does WolframAlpha. I found the problem and its solution here: https://brilliant.org/wiki/squeeze-theorem/

Where am I wrong????!!!!

Thanks for any answers!

thornsword
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  • Limit of sum is sum of limits is not true for infinite sums like yours. To see this, look at the infinite sequence 1, -1, repeating. It is undefined but $\sum \lim (1-1) = 0$. – Hubble Apr 08 '20 at 15:26
  • The addition property of limits can be applied to a FIXED number of terms. Here the number of terms is $n$ which varies as $n\to\infty$. – Robert Z Apr 08 '20 at 15:31
  • But why can't we extend the proof for finite terms to infinite terms? In case of 2 terms, we use the usual $\frac{\epsilon}{2}$ approach...why can't we just extend it to be true for $n$ terms? (and then take the limit as $n\to \infty?$) – thornsword Apr 08 '20 at 15:33

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