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$\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+1}+\dfrac{n}{n^2+2}+\dfrac{n}{n^2+3}\cdots\cdots+\dfrac{n}{n^2+n}\right)$

Can we write it as following

$E=\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+1}\right)+\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+2}\right)+\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+3}\right)\cdots\cdots+\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+n}\right)\tag{1}$

Let's see what happens:-

$$\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+1}\right)$$ $$\lim\limits_{n\to\infty}\dfrac{\dfrac{1}{n}}{1+\dfrac{1}{n^2}}=0$$

In the same way for further terms, we will get $0$

Let's also confirm for general term

$$\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+n}\right)$$ $$\lim\limits_{n\to\infty}\left(\dfrac{\dfrac{1}{n}}{1+\dfrac{1}{n}}\right)=0$$

So the whole expression $E$ will be zero

But actual answer is $1$

Let's see what happens if we evaluate the original expression $OE=\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+1}+\dfrac{n}{n^2+2}+\dfrac{n}{n^2+3}\cdots\cdots+\dfrac{n}{n^2+n}\right)$

$OE=\lim\limits_{n\to\infty}\left(\dfrac{\dfrac{1}{n}}{1+\dfrac{1}{n^2}}+\dfrac{\dfrac{1}{n}}{1+\dfrac{2}{n^2}}+\dfrac{\dfrac{1}{n}}{1+\dfrac{3}{n^2}}\cdots\cdots+\dfrac{\dfrac{1}{n}}{1+\dfrac{1}{n}}\right)$

Now we can easily see that each term inside the bracket is tending to $0$, so can we say that sum of all terms upto infinity as well tends to zero?

I think we cannot because the quantity is not exactly zero, it is tending to zero, so when we add the values tending to zero upto infinity, we may not get zero.

But I got the following counter thought:-

$\lim\limits_{x\to0}\dfrac{(1+x)^\frac{1}{3}-1}{x}$

As we know $(1+x)^n=1+nx+\dfrac{n(n-1)}{2}x^2+\dfrac{n(n-1)(n-2)}{6}x^3\cdots\cdots\infty$ where $|x|<1$

$\lim\limits_{x\to0}\dfrac{\left(1+\dfrac{1}{3}x-\dfrac{1}{3}\cdot\dfrac{2}{3}\cdot\dfrac{1}{2}x^2+\dfrac{1}{3}\cdot\dfrac{2}{3}\cdot\dfrac{5}{3}\cdot\dfrac{1}{6}x^3\cdots\cdots\right)-1}{x}$

$\lim\limits_{x\to0}\dfrac{1}{3}-\dfrac{1}{3}\cdot\dfrac{2}{3}\cdot\dfrac{1}{2}x+\dfrac{1}{3}\cdot\dfrac{2}{3}\cdot\dfrac{5}{3}\cdot\dfrac{1}{6}x^2\cdots\cdots$

Now here also all the terms except $\dfrac{1}{3}$ are tending to $0$. So here also we can say that the whole quantity may not turn out to be zero as we are adding all terms upto infinity.

But surprisingly $\dfrac{1}{3}$ is the correct answer.

I am feeling very confused in these two things. Please help me.

Martin Sleziak
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prat
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7 Answers7

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Observe that $$\frac{n}{n^2+1}+\frac{n}{n^2+2}+\cdots+\frac{n}{n^2+n}$$ lies between $$\frac{n}{n^2+n}+\frac{n}{n^2+n}+\cdots+\frac{n}{n^2+n}=\frac{n}{n+1} $$ and $$\frac{n}{n^2}+\frac{n}{n^2}+\cdots+\frac{n}{n^2}=1.$$

Angina Seng
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I will be giving a $\varepsilon -\mathcal{N}$ proof

notice

$$\begin{align}\frac{n}{1+n^2} + \frac{n}{2+n^2} + \cdots + \frac{n}{n+n^2} - 1= n\Big( \frac{1}{1+n^2} - \frac{1}{n^2} + \cdots + \frac{1}{n+n^2} - \frac{1}{n^2} \Big) = h_n\end{align}$$

Claim $\dfrac{n}{n+n^2} > \dfrac{r}{r+n^2}$

So

$|h_n| < \Big|n\cdot(\dfrac{1}{n^2}\cdot\dfrac{n^2}{n+n^2})\Big|=\dfrac{1}{n+1}<\dfrac{1}{n}$

so $\forall n > \dfrac{1}{\varepsilon}$ $|h_n| < \varepsilon$

Aditya Dwivedi
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That is not legitimate, because the number of terms inside the limit are growing as $n$ tends to infinity.

Rather, one sees that \begin{align*} \sum_{k=1}^{n}\dfrac{n}{n^{2}+n}\leq\sum_{k=1}^{n}\dfrac{n}{n^{2}+k}\leq\sum_{k=1}^{n}\dfrac{n}{n^{2}+1}, \end{align*} the left and right-sided both tend to $1$.

user284331
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  • Ok then what about the second case – prat Dec 08 '19 at 03:19
  • The number of terms do not grow as $x$ is running. – user284331 Dec 08 '19 at 03:20
  • but if you see the formula carefully, it goes till infinity – prat Dec 08 '19 at 03:21
  • Yes, but the terms are still that much. If you see the first one, each times $n$ varies, there are $n$ terms going on, but the second case no, there is such an amount (assume you can really count them) no matter what $x$ is, both are in different circumstances. – user284331 Dec 08 '19 at 03:23
  • ok well said, but then what is the answer for this $\lim\limits_{x\to0}\dfrac{(1+x+x^2+x^3+x^4\cdots\cdots\infty)-1}{x}$? – prat Dec 08 '19 at 03:26
  • The limit is about $\dfrac{\dfrac{1}{1-x}-1}{x}=\dfrac{1}{1-x}\rightarrow 1$. – user284331 Dec 08 '19 at 03:28
  • ok, your way is correct. Suppose I try to solve it in this way:

    $\lim\limits_{x\to0}\dfrac{x+x^2+x^3+x^4\cdots\cdots\infty}{x}$

    $\lim\limits_{x\to0}1+x+x^2+x^3\cdots\cdots\infty$

    Now If I just say that as $x\rightarrow 0$ so $1+x+x^2+x^3\cdots\cdots\infty$ tends to one. But this is the wrong way as previously discussed but I am getting the same answer.

     – prat Dec 08 '19 at 03:33
  • This is correct, and again the amount doesn't grow as $x$ is going. The amount is like the amount of all natural numbers, but fine, the amount is fixed. – user284331 Dec 08 '19 at 03:34
  • but in the question posted, how the amount is growing over there, there also every term is roughly zero – prat Dec 08 '19 at 03:38
  • By my wording of amount growing, I didn't mean that it can't be infinite amount of terms, I meant, rather, the amount is fixed, in the first one, say, when $n=5$, there are five terms, when $n=100$, there are hundreds, but in the second one, no matter what $x$ is, the amount is natural number many, still, natural number many, so to speak. – user284331 Dec 08 '19 at 03:40
  • ok, so in the original post, denominator is decreasing term by term and numerator is same. So that's why even though terms are roughly zero but its a divergent series and we can't sum in the trivial answer. Is this correct? – prat Dec 08 '19 at 03:46
  • I don't think this is the true reason. But rather, summing infinitely many terms need to be careful, for the $x$ thing, there is a theory about power series why you can do that, in general it is hard to say what reason it is. – user284331 Dec 08 '19 at 03:48
  • now I am confused , can you please tell me how you figured it out that the amount is growing and not growing in other cases. Can you please tell me step by step. – prat Dec 08 '19 at 03:50
  • @prat: You need to actually look carefully at the theorem you are attempting to use to split the limit. Don't ignore all the conditions! If it says "If X and Y exist then ... exist, then make sure you check that you only apply it to two objects that exist". For example, it is true that "given any two natural numbers $m,n$, their sum $m+n$ is also a natural number", but it is false that "$1+2+3+\cdots$ is a natural number". In fact, "$1+2+3+\cdots$" is even meaningless, because no theorem permits you to write it down as a meaningful expression! – user21820 Dec 08 '19 at 13:14
  • That said, I'm happy to see that you put in a lot of effort into your question, so I don't think it deserves any downvotes. – user21820 Dec 08 '19 at 13:15
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If you know harmonic numbers $$S_n=\sum_{i=i}^n \frac 1 {n^2+i}=H_{n^2+n}-H_{n^2}$$ Using the asympotics $$H_p=\gamma +\log \left({p}\right)+\frac{1}{2 p}-\frac{1}{12 p^2}+O\left(\frac{1}{p^3}\right)$$ apply it twice and continue with Taylor series to get $$n S_n=1-\frac{1}{2 n}-\frac{1}{6 n^2}+O\left(\frac{1}{n^3}\right)$$ which shows the limit nd how it is approached.

Moreover, this gives a quite good approximation of the sum. Using $n=10$, the exact value is $\frac{11210403701434961}{11818204429243212} \approx 0.94857$ while the above truncated series gives $\frac{569}{600}\approx 0.94833$.

Claude Leibovici
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You can use the Squeeze Theorem, with

$$ n\cdot \frac{n}{n^2 + n} \leq a_n \leq n\cdot \frac{n}{n^2 + 1} $$

And as $ n \to \infty$, you get $ 1 \leq \lim_{n \to \infty} a_n \leq 1 $, So, using the Squeeze Theorem, $a_n \to 1$.

talbi
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  • But $\lim_{n\to\infty}an$ for n/n+1 sequence is 1(not 0). So will the series n/n+1 converge then? – Abhinav Tahlani Dec 06 '20 at 14:41
  • I'm not sure what you mean, I've never wrote that the limit of something here is $0$. $\frac{n}{n+1}$ does converge to $1$. – talbi Dec 06 '20 at 14:47
  • I mean that if a series has to converge, its nth term(when n is tending to infinity) has to tend to zero. But for your RHS, that isn't happening. – Abhinav Tahlani Dec 06 '20 at 14:51
  • On the RHS, we have $n \cdot \frac{n}{n^2+1} = \frac{n^2}{n^2+1}$. This converges to $1$, not zero! Perhaps you missed the multiplication by $n$? – talbi Dec 06 '20 at 14:54
  • My bad, LHS I meant! – Abhinav Tahlani Dec 06 '20 at 15:07
  • The same applies to LHS, $\frac{n^2}{n^2+n}$. You can divide by $n^2$ and get $\frac{1}{1+1/n}$, which approaches $1$ as $ n \to \infty$. – talbi Dec 06 '20 at 15:17
  • Yeah I wanted to say that only. The limit of the sequence approaches 1, not zero. So how can the series converge? – Abhinav Tahlani Dec 06 '20 at 15:43
  • I probably should've said, but $a_n$ refers to the sum, not the individual term. So, $a_n = \frac{n}{n^2+1} + \frac{n}{n^2 + 2} + \ldots + \frac{n}{n^2+n}$. You are right, the individual term approaches $0$, but the sum approaches $1$. The series ($\frac{n}{n^2 + k}$) converges since there's a 2nd degree polynomial on the denominator – talbi Dec 06 '20 at 16:32
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{\bbox[5px,#ffd]{}}$

\begin{align} &\bbox[5px,#ffd]{\sum_{k = 1}^{n}{n \over n^{2} + k}} = n\sum_{k = 0}^{n - 1}{1 \over k + 1 + n^{2}} \\[5mm] = &\ n\sum_{k = 0}^{\infty}\pars{{1 \over k + 1 + n^{2}} - {1 \over k + n + 1 + n^{2}}} = n\pars{H_{n^{2} + n} - H_{n^{2}}} \\[5mm] \stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\,\,\,& n\braces{\bracks{\ln\pars{n^{2} + n} + \gamma + {1 \over 2n^{2} + 2n}} - \bracks{\ln\pars{n^{2}} + \gamma + {1 \over 2n^{2}}}} \\[5mm] \stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\,\,\, &\ n\ln\pars{1 + {1 \over n}} \stackrel{\mrm{as}\ n\ \to\ \infty}{\to}\,\,\, \bbx{1} \\ & \end{align} $\ds{H_{z}}$ is a Harmonic Number and $\ds{\gamma}$ is the Euler-Mascheroni Constant.
Felix Marin
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Following the logic in your answer $$ \lim_{n\to\infty}\overbrace{\left(\frac1n+\frac1n+\cdots+\frac1n\right)}^\text{$n$ terms}=0 $$ since each term tends to $0$. This is obviously false since the sum for each $n$ is $1$, so the limit is $1$.

We can say that a finite sum of terms which have a limit equals the finite sum of the limits. However, the same cannot be said for an unlimited sum of terms even though each has a limit. We must have some other theorem that allows us to swap order of the limits.

One proper way to handle the sum in the question is to bound the sum with $$ \frac{n}{n+1}\le\left(\frac{n}{n^2+1}+\frac{n}{n^2+2}+\cdots+\frac{n}{n^2+n}\right)\le1 $$ and then the Squeeze Theorem says the limit is $1$.

robjohn
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  • I think something wrong is there in your first statement, it should be $OE=\lim\limits_{n\to\infty}\left(\dfrac{\dfrac{1}{n}}{1+\dfrac{1}{n^2}}+\dfrac{\dfrac{1}{n}}{1+\dfrac{2}{n^2}}+\dfrac{\dfrac{1}{n}}{1+\dfrac{3}{n^2}}\cdots\cdots+\dfrac{\dfrac{1}{n}}{1+\dfrac{1}{n}}\right)$ and here I don't see that it is obviously $1$? Please correct me if I am wrong. – prat Dec 08 '19 at 04:53
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    @prat: No, the first statement is a simpler example that shows why the logic used in your solution fails. The second statement shows how to see that the limit of the sum in the question is $1$ by applying the Squeeze Theorem. – robjohn Dec 08 '19 at 05:04
  • yeah you are right, I only have question,

    What happened in the second case mentioned in the question, why over there same logic doesn't apply?

     – prat Dec 08 '19 at 05:10
  • The sum is $$\sum_{n=0}^\infty\frac{\frac43-1}1\frac{\frac43-2}2\cdots\frac{\frac43-n}nx^n$$ the terms of which are dominated by $|x|^n$, so we can use Dominated Convergence to say the series will converge for $|x|\lt1$. – robjohn Dec 08 '19 at 07:06