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Show that:-
$$\lim_{x \rightarrow 0 } \frac{1}{6x^2} - \frac{1}{120x^4} +.... ({-1})^{n+1}\frac{1}{(2n+1)!x^{2n}}$$ for $n\geq1$ where n $\in$ $\mathbb{Z^+}$ is equal to $1$ as $n\to\infty$.

My method was as the form of the limit is $\infty - \infty$ , so i tried first assuming limit exists and taking common denominator for a finite n and tried with lhopital if the limit exists but it was not working , may anyone tell ?

DatBoi
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Paracetamol
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  • Just to confirm, are you trying to compute the following: $$\lim_{n\to\infty}\lim_{x\to 0}\sum_{i=1}^n(-1)^{i+1}\frac{1}{(2i+1)x^{2i}}$$ If this is what you are trying to solve then use Stolz-Cesaro (https://en.wikipedia.org/wiki/Stolz%E2%80%93Ces%C3%A0ro_theorem) – Marcos May 10 '22 at 12:59
  • Yeah right @Marcos okay will see it – Paracetamol May 10 '22 at 13:02
  • If you want to interchange limits, then see https://math.stackexchange.com/questions/15240/when-can-you-switch-the-order-of-limits – Marcos May 10 '22 at 13:03
  • Hint Divide by $x$ and make $x=\frac 1y$ and you are very close to a well know series – Claude Leibovici May 10 '22 at 13:42

1 Answers1

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$$S=\sum_{n=1}^{\infty}({-1})^{n+1}\frac{1}{(2n+1)!x^{2n}}$$

$$\frac{e^{x}-1-x}{x}=\sum_{n=1}^{\infty}\frac{x^{n}}{(n+1)!} $$

$$\frac{e^{1/x}-1-\frac1x}{\frac1x}=\sum_{n=1}^{\infty}\frac{1}{(n+1)!x^{n}}$$

$$\frac{e^{1/x}-1-\frac1x}{\frac1{x}}+\frac{e^{1/-x}-1+\frac1x}{\frac1{-x}}=2\sum_{n=1}^{\infty}\frac{1}{(2n+1)!x^{2n}}$$

$$x({e^{1/x}}-{e^{1/-x}}-\frac2x)=2\sum_{n=1}^{\infty}\frac{1}{(2n+1)!x^{2n}}$$

Substituting $ x \to ix$

$$ix({e^{1/ix}}-{e^{1/-ix}}-\frac{2}{ix})=-2\sum_{n=1}^{\infty}\frac{({-1})^{n+1}}{(2n+1)!x^{2n}}$$

$$S=-\frac{ix}2(e^{-i/x}-e^{i/x}+\frac{2i}x)$$

$$\lim_{x\to0}S=1$$

Clemens Bartholdy
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DatBoi
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  • Where do you justify interchanging the limits? The question wants $x\to0$ first, and then $n\to\infty$. – Lorago May 10 '22 at 14:19
  • @Lorago my knowledge in mathematics is way too little to comprehend the complexities of the monotone convergence theorem. In case the interchange is valid, I'll leave this as a reference to future readers, despite the downvotes. – DatBoi May 10 '22 at 14:23
  • Added the substitution text. I guess one issue is that when you substitued $ x \to ix$ , the amount of mathematical theory to describe what you've done formally increased quite a bit because you're then dealing with a complex limit. In the complex plane, you're essentially approaching the origin along the line $y=0$ (see how the ix expression varies). – Clemens Bartholdy May 17 '22 at 09:34
  • This a quite a bit complex method, i didnt understood clearly but i am fine with dealing with this later so as to get what it fully means – Paracetamol May 17 '22 at 11:31
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    @Aplateofmomos I simply tried to manipulate the taylor expansion to the desired form and I'm surprised to see that there are so many nuances to it(multivariable limit, limit interchange, etc.). Please forgive my ignorance :) – DatBoi May 17 '22 at 13:17
  • I understand, don't worry :) @DatBoi – Clemens Bartholdy May 17 '22 at 14:01