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Find the limit of $\lim_{x\to \infty}(\frac{x}{x})^x+(\frac{x-1}{x})^x+(\frac{x-2}{x})^x......+(\frac{1}{x})^x$

$\lim_{x\to \infty}(\frac{x}{x})^x+(\frac{x-1}{x})^x+(\frac{x-2}{x})^x......+(\frac{1}{x})^x$

$=\lim_{x\to \infty}\frac{1^x+2^x+3^x+.....+x^x}{x^x}$
This is in $\frac{\infty}{\infty}$ form ,so i applied L hospital rule.
$=\lim_{x\to \infty}\frac{2^x\log 2+3^x\log 3+.....+x^x(1+\log x)}{x^x(1+\log x)}$
But i am stuck here and could not solve further.Please help.

diya
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    Does $x$ run through the natural numbers? – sranthrop Dec 13 '15 at 04:09
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    By the way, you can't take the derivative of a sum with variable number of terms that way. Just consider the function $x^2=x+x+x+\dots+x$. The derivative your way would yield $1+1+1+\dots+1=x$. Also, the function isn't actually defined except for $x$ an integer, so it does not satisfy the most basic need for having a derivative. You'd have to come up with some natural continuous version of the numerator for this to work. – Thomas Andrews Dec 13 '15 at 06:26
  • That would all be more clear if the problem hadn't chosen $x$ as the variable. We often assume a variable named $x$ is continuous, while variables named $n$ or $m$ are integers. – Thomas Andrews Dec 13 '15 at 06:28

4 Answers4

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For a fixed $m$:

$$\lim_{x\to\infty}\left(\frac{x-m}{x}\right)^x= e^{-m}\tag{1}$$

So if the series converges at all, it must converge to a value at least as big as:

$$\sum_{m=0}^\infty e^{-m} = \frac{1}{1-e^{-1}}=\frac{e}{e-1}$$

Letting $f_m(x)=\left(\frac{x-m}{x}\right)^x$, then define $g_m(x)=\ln f_m(x)=x\left(\log(x-m)-\log(x)\right)$, and you get:

$$g_n'(x)=\log(x-m)-\log(x) + \frac{x}{x-m} - 1=\log\left(1-\frac{m}{x}\right) +\dfrac{\frac{m}{x}}{1-\frac mx}$$

You can use the power series for $\frac{1}{1-z}$ and $\log(1-z)$ to show that this value is positive, and hence $g_m$ is increasing for $x>m$, and this that the above series converges to the hoped-for value.

This shows that $F(x)=\sum_{0}^x f_m(x)$ is strictly increasing, and thus we have:

$$\lim_x F(x)=\sup_x F(x)\leq \sum_{m} e^{-m}$$ And also have shown above that $$\sup_x F(x)\geq \sum_{m} e^{-m}$$

So the limit is equal to $\frac{e}{e-1}$.

Aside: If you have a sequence of sequences, $a_{n,k}$, and for each $k$, you have $\lim_{n\to\infty} a_{n,k}=A_k$, it is not always true that:

$$\lim_{n\to \infty}\sum_{k=0}^\infty a_{n,k} = \sum_{k=0}^\infty A_k$$

A simple example where it isn't true is: $$a_{n,k}=\delta_{n,k}=\begin{cases}1&n=k\\0&n\neq k\end{cases}$$

Then for each $k$, $\lim_{n\to\infty} a_{n,k}=0$, and for all $n$, $\sum_{k=0}^{\infty}a_{n,k}=1$, so the limit of the sum is not the sum of the the limits.

If, for each $k$, however, $a_{1,k},a_{2,k},\dots$ is increasing, then the limit of the sums is the sum of the limit.

Thomas Andrews
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  • See my answer for a proof of the general theorem. =) – user21820 Dec 13 '15 at 06:15
  • The series is monotonically increasing (Bernoulli's Inequality) and so the limit of the sum is the sum of the limits (Monotone Convergence). – robjohn Dec 13 '15 at 11:39
  • Why the series converges to at least $\sum_{m=0}^\infty e^{-m}$, if it converges? – robit Dec 14 '15 at 06:36
  • @robit: See my answer. In particular, it crucially depends on the non-negativity. – user21820 Dec 14 '15 at 06:51
  • After reading your answer, I see that the series converges to at least $\sum_{m=0}^l e^{-m}$ for $l \in \mathbb{N}$ but not $\sum_{m=0}^\infty e^{-m}$ if it converges, which is a finite summation. @user21820 – robit Dec 14 '15 at 07:39
  • @robit: That's only the first part of that part of my answer. The second part shows that we can strengthen it to the infinite sum simply by going backwards; choose a sufficiently long initial segment of the infinite sum, which is a finite sum, and hence the series converges to at least a little less than the infinite sum. – user21820 Dec 14 '15 at 10:10
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Note that $$ \begin{align} \lim_{x\to\infty}\left(\frac{x-k}x\right)^x &=\lim_{x\to\infty}\left(1-\frac kx\right)^x\\ &=e^{-k} \end{align} $$ Therefore, the sum of the limits is $$ 1+e^{-1}+e^{-2}+\dots=\frac1{1-\frac1e}=\frac e{e-1} $$

First, since the series was given as $\left(\frac xx\right)^x+\cdots+\left(\frac1x\right)^x$, I assumed that $x\in\mathbb{Z}$.

However, even if we don't assume that $x\in\mathbb{Z}$, but that we only include $\left(\frac{x-k}x\right)^x$ for $k\lt x$, Bernoulli's Inequality says that $\left(1-\frac kx\right)^x$ is increasing in $x$. That is, for $y\ge x\ge k$, $$ \overbrace{\left(1-\frac ky\right)^{\large\frac yx}\ge1-\frac yx\frac ky}^{\text{Bernoulli's Inequality}} \implies \overbrace{\left(1-\frac ky\right)^y\ge\left(1-\frac kx\right)^{x\vphantom{\large\frac yx}}}^{\text{raising to the $x$ power}} $$ Thus, we only need invoke Monotone Convergence (the terms for $k\ge x$ are $0$) to assure that the limit of the sum is the sum of the limits.

robjohn
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    That assumes a niceness of convergence. Try proving the general theorem about a sequence of series, and you'll see you need a little more. – Thomas Andrews Dec 13 '15 at 04:33
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    For example, $a_{n,k}=\delta_{n,k}$ has the property that $\lim_k a_{n,k}=0$ for all $n$, but $\lim_{n} \sum_k a_{n,k}=1$. – Thomas Andrews Dec 13 '15 at 04:34
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    I frankly do not understand why you gave such a misleading answer, and Thomas has already pointed out the serious conceptual flaw. I would like to assume that you know how to do it correctly, but most students don't. @diya: I do not understand why you accepted this answer despite Thomas' answer mentioning the subtle issue. – user21820 Dec 13 '15 at 04:46
  • Likely just went to bed after writing the answer, so didn't seem my comments. @user21820 – Thomas Andrews Dec 13 '15 at 06:10
  • My answer was not wrong, just lacking a mention of Bernoulli's Inequality and Monotone Convergence. Sorry for the omission, but I was called away as I was writing this and did not get back until this morning. Since the problem the OP had was their approach, I concentrated on that first. – robjohn Dec 13 '15 at 11:33
  • @user21820: I did not see Thomas Andrews' answer until I came back to the site. As soon as I did, I added the part about Bernoulli's Inequality and Monotone Convergence. Now that I read your answer, you mention Monotone Convergence, however Bernoulli's Inequality simplifies things further. – robjohn Dec 13 '15 at 11:54
  • Not really seeing how Bernoulli proves increasing. – Thomas Andrews Dec 13 '15 at 13:20
  • @ThomasAndrews: I have changed $\left(\frac{x-k}{x}\right)^x$ to $\left(1-\frac kx\right)^x$ hopefully to make it clearer. If not, then Bernoulli's Inequality says that for $y\ge x\ge k$ $$\left(1-\frac ky\right)^{\frac yx}\ge1-\frac yx\frac ky$$ raise both sides to the $x$ power to get $$\left(1-\frac ky\right)^y\ge\left(1-\frac kx\right)^x$$ – robjohn Dec 13 '15 at 13:36
  • Yeah, that was the step I was missing - $(x+y)/x$. – Thomas Andrews Dec 13 '15 at 13:38
  • I did at first want to say that the monotonicity can be proven via Bernoulli's inequality, but I decided that it was better to have a fuller understanding of the behaviour, which can be seen via the asymptotic expansion. As for omissions, I consider it wrong as long as it is a gap that students cannot fill in themselves, because they would be misled to think that the reasoning is right. Countless times we see questions on Math SE due to an utterly broken concept of limit manipulations. Worse still, some teachers insist on teaching error (http://matheducators.stackexchange.com/a/9917/1550)!!! – user21820 Dec 14 '15 at 05:37
  • As I said, I first addressed the OP's main problem, which was their approach. I felt that if they noticed $\left(1-\frac kx\right)^x\to e^{-k}$, that would help (they asked for help). Unfortunately, I had to leave for the evening before I could write a complete answer, when I came back, I saw Thomas's comment and completed the answer. It would probably have been better to have prefaced that the initial answer was a hint, or that the limit was monotonic before I left, but I had to leave quickly so I submitted what I had. – robjohn Dec 14 '15 at 05:58
  • Yes I understand your point, and that's why I edited the preface to my answer. It's simply that the limit you mention is nowhere near as crucial to this problem as the monotone convergence theorem, and so most students who read your original answer would be misled in the way I've stated. In fact, it should have been clear from the asker's question itself that he/she didn't know what he/she was doing, what with differentiating a sum of $x$ terms with respect to $x$. So I stand by my judgement that it is incorrect to omit any of the conceptual points in a solution. – user21820 Dec 14 '15 at 06:20
  • I have changed the original statement to "the sum of the limits is" which is correct. The second part shows that the limit of the sum is the sum of the limits – robjohn Dec 14 '15 at 07:03
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Edit: robjohn has corrected his answer, but I'll leave my proof of the general theorem here, since the techniques involved are instructive. $\def\nn{\mathbb{N}}$

Solution

In general the following MCT (monotone convergence theorem) holds:

$\sum_{k=0}^\infty f_n(k) \to \sum_{k=0}^\infty f(k)$ as $n \to \infty$ for any increasing sequence of non-negative functions $(f_n)_{n\in\nn}$ such that $f_n \to f$ pointwise as $n \to \infty$.

In this particular instance $f_n(k) = (1-\frac{k}{n})^n$ for any $n \in \nn$ and $k \in \nn_{\le n}$, and $f_n(k) = 0$ for any $n \in \nn$ and $k \in \nn_{>n}$. To apply MCT you must first prove that $(1-\frac{k}{n})^n \le (1-\frac{k}{n+1})^{n+1}$ for any $n,k \in \nn$, which is easy to see from $(1-\frac{k}{n})^n = \exp(n\ln(1-\frac{k}{n})) = \exp( - \sum_{i=1}^\infty \frac{k^i}{n^{i-1}} )$. Then MCT gives immediately that $\sum_{k=0}^{n-1} (1-\frac{k}{n})^n \to \sum_{k=0}^\infty e^{-k} = \frac{1}{1-e^{-1}}$.

Proof of MCT

Clearly $\sum_{k=0}^\infty f_n(k) \le \sum_{k=0}^\infty f(k)$ for any $n \in \nn$, because $f_n(k) \le f_{n+1}(k)$ for any $n,k \in \nn$.

On the other hand, for any $c \in \nn$, we have $\sum_{k=0}^\infty f_n(k) \ge \sum_{k=0}^c f_n(k) \to \sum_{k=0}^c f(k)$ as $n \to \infty$, where the inequality is by non-negativity of $(f_n)_{n\in\nn}$, and the limit is by their pointwise convergence to $f$ and because we only have a limit of a finite sum. But $\sum_{k=0}^c f(k) \to \sum_{k=0}^\infty f(k)$ by definition of infinite series, where the infinite sum exists (could be $\infty$) by non-negativity. Thus given any $ε > 0$, we can choose $c \in \nn$ such that $\sum_{k=0}^c f(k) > \sum_{k=0}^\infty f(k) - ε$, and then choose $m \in \nn$ such that $\sum_{k=0}^c f_n(k) > \sum_{k=0}^c f(k) - ε$ for any $n \in \nn_{\ge m}$. This implies that $\sum_{k=0}^c f_n(k) > \sum_{k=0}^\infty f(k) - 2ε$ for any $n \in \nn_{\ge m}$.

Together the above two facts imply that $\sum_{k=0}^c f_n(k) \to \sum_{k=0}^\infty f(k)$ as $n \to \infty$.

Notes

It is instructive to see why each of the conditions for MCT is important. It is also related to corresponding theorems in measure theory such as MCT for sets or MCT for functions.

user21820
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When x tends to infinity, the first term containing x/x is obviously equal to one. Rest all terms are of the form, "something between zero and 1 raised to power infinity". So all other terms are going to be zero. Therefore answer should be just '1'.

Asif Kazmi
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