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$$\lim_{k\to\infty} \left[\frac{e^{1/k}}{k^2}+\frac{2(e^{1/k})^2}{k^2}+\frac{3(e^{1/k})^3}{k^2}+ \dots +\frac{k(e^{1/k})^k}{k^2}\right]$$

(Here is a picture of the problem)

I have doubt in that question (where he has used $e^{1/k}$) My doubt is as $k \to \infty$, $1\over k$ should tend to zero therefore $e^{1/k } \to 1$, so ${e^{1/k}\over k^2}$ should tend to zero because denominator is very huge so all series will end to zero but this is not the answer, the answer is 1. My teacher has done it by AGP method with which he got answer 1 please explain..

Martin Sleziak
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user5954246
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    Are you sure that this is a limit and not a series? – Morgormir Feb 18 '16 at 19:10
  • hey i added a picture of question are u able to see – user5954246 Feb 18 '16 at 19:11
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    The question in the image is not asking you to find the limit of a single term of the summation, but rather to find the limit of the sum of the terms in the expression. You may not push the limit inside of the summation, evaluate the limit first, and sum the result. Limit of sum of terms is not sum of limits of terms. – JMoravitz Feb 18 '16 at 19:13
  • hey but according to sum of limit property limit of sums is equal to sum of limits – user5954246 Feb 18 '16 at 19:14
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    That is not true in general, especially when the number of terms being added depends on the variable that changes in size. Consider the following: $1 = \frac{1}{2}+\frac{1}{2}=\frac{1}{3}+\frac{1}{3}+\frac{1}{3}=\dots = (\underbrace{\frac{1}{n}+\dots+\frac{1}{n}}_{n~\text{times}})\neq (0+0+\dots+0)=0$ – JMoravitz Feb 18 '16 at 19:16
  • Finite sums, not infinite sums, consider $\lim_{k \to \infty} \sum_{k=0}^n (-1)^k$ – Morgormir Feb 18 '16 at 19:17
  • oh!!! i think i got a bit idea – user5954246 Feb 18 '16 at 19:20
  • hey but i am substituting the value of k everywhere so the limit should exist ! like limit(x tends to 0) x^2+x ..... when i am substituting limit everywhere i get correct answer but why not here – user5954246 Feb 19 '16 at 11:19
  • What is happening to some degree is that you are looking at $\lim_{x\to c} f(x,g(x))$ and you are trying to say that it should be $\lim_{x\to c} f(x,g(x))= f(x,\lim_{x\to c} g(x))$ which is incorrect. The actual limit you are trying to calculate depends on more than just the terms being added. Consider my earlier counterexample. $1 = \lim_{n\to\infty} \sum\limits_{k=1}^n \frac{1}{n} \neq \sum\limits_{k=1}^{ \color{red}{n}} \lim_{n\to\infty} \frac{1}{n} = 0$. You essentially pushed the limit past some of the other terms you were wishing to change, the red $n$ in this case. – JMoravitz Feb 19 '16 at 12:54
  • Your summation: $\lim_{k\to\infty} \sum\limits_{i=1}^k \frac{i(e^{1/k})^i}{k^2} \neq \sum\limits_{i=1}^{\color{red}{k}} \lim_{k\to\infty} \frac{i(e^{1/k})^i}{k^2}$ – JMoravitz Feb 19 '16 at 12:58
  • What is AGP method? – Martin Sleziak Feb 22 '16 at 16:59
  • Maybe you could first simply $e^{1/k}+2(e^{1/k})^2+\dots+k(e^{1/k})^k$ (you can use the formula derived in this post). And then use limit on the result. – Martin Sleziak Feb 22 '16 at 17:10

1 Answers1

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It is true that $\lim_{k \to \infty}\frac {e^{1/k}}{k^2}=0$ for the reason you stated. Either your teacher is wrong or you have misunderstood.

Martin Sleziak
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Ross Millikan
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