3

I compared the method used for the following 2 questions:

(i) $$\lim_{x\to 0}\frac{x-\sin x}{x^3}=\lim_{x\to 0}\frac{x-\left(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\ldots\right)}{x^3}=\lim_{x\to 0}\left(\frac{1}{3!}-\frac{x^2}{5!}+\frac{x^4}{7!}-\ldots\right)=\lim_{x\to 0}\frac{1}{3!}-\lim_{x\to 0}\frac{x^2}{5!}+\lim_{x\to 0}\frac{x^4}{7!}-\ldots=\frac{1}{3!}$$

(ii)$$\lim_{n\to \infty}\frac{1+2+3+\ldots+n}{n^2}=\lim_{n\to \infty}\left(\frac{1}{n^2}+\frac{2}{n^2}+\frac{3}{n^2}+\ldots+\frac{n}{n^2}\right)=\lim_{n\to \infty}\frac{1}{n^2}+\lim_{n\to \infty}\frac{2}{n^2}+\lim_{n\to \infty}\frac{3}{n^2}+\ldots+\lim_{n\to \infty}\frac{n}{n^2} =0$$

The second is wrong as we know that the limit is $\frac{1}{2}$. Is the first method acceptable? What is condition for $$\lim_{n\to a}\sum_{k=1}^\infty f_k(n)=\sum_{k=1}^\infty \lim_{n\to a} f_k(n)?$$

pipi
  • 2,471
  • Consider $f$ and $g$ with $n$-th order Taylor polynomials $P(x)$ and $Q(x)$ about $a$, and remainders $R$ and $S$ respectively. Suppose that $\lim_{x \to a} \frac{P(x)}{Q(x)}$ exists; then $\lim_{x \to a} \frac{P(x)}{Q(x)} = \lim_{x \to a} \frac{P(x) + R}{Q(x)}$ (since R tends to zero over $(x-a)^n$) = $\lim_{x \to a} \frac{P(x) + R}{Q(x)} * \lim_{x \to a}\frac{Q(x)}{Q(x) + S}$ (for the same reason). But this last limit is $\lim_{x \to a} \frac{f}{g}$. – Chris Jan 16 '13 at 02:20
  • I.e., there's a reason why you can do what you did in the first case! Specifically, instead of considering the whole series, just consider the 3rd order Taylor polynomials. – Chris Jan 16 '13 at 02:22
  • You lost "infinite $n$" in your second example! – RSh Aug 18 '13 at 02:59

2 Answers2

7

The equality $$\lim_{n\to a}\sum_{k=1}^\infty f_k(n)=\sum_{k=1}^\infty \lim_{n\to a} f_k(n)$$ holds under the condition of uniform convergence of the series with respect to the parameter $n$. Uniform convergence means: for every $\epsilon>0$ there is $K$ such that $$\left|\sum_{k=K}^\infty f_k(n)\right|<\epsilon$$ for all $n$ in some fixed interval containing $a$. The uniformity is in the fact that $K$ is chosen independently of $n$.

For example, the Taylor series for sine converges uniformly if $x$ is restricted to any bounded interval, such as $[-1,1]$.

Your second example is not written in the form $\lim_{n\to a}\sum_{k=1}^\infty f_k(n)$ since the number of summands is finite and depends on $n$. You could rewrite it as such, by using zeros for missing terms. But the convergence is not uniform. No matter how large $K$ we take, if $n>2K$, the tail sum $$\sum_{k=K}^{2n}\frac{k}{n^2}>\sum_{k=K}^{2n}\frac{n/2}{n^2}=\frac12$$ is not small.

user
  • 1,062
0

I do notice that in the first expression each term involves a function of the limiting variable: x.

Whereas in the second expression each terms involves a constant number up to the value of the limiting variable: x.

The first Sum is invariant to the value of its limiting variable, it goes to infinity.

The second Sum is dependent on its value of its limiting variable, it goes until the value of the limiting variable.

That I think is the issue here. Because when you are evaluating the second limit, the number of terms you have is CHANGING along with the value of the variable as it approaches infinity. Depending on what value you give N the number of terms is roughly equal to N. You also have the issue that your current formula isn't well defined for non Integral N. Whereas the first expression is always defined for Real (and as a matter of fact Complex) N. These two properties combined might have an effect on evaluating the expression. Because the idea of a limit isn't to just evaluate at a point but to "slide" or "limit" towards it.

In order to resolve it (which you probably know) you can rewrite 1 + 2 + 3 ... N as 1/2 * N *(N + 1) and then take the limit to infinity to get 1/2.

How to know when to do this in more obscure sums is really tricky to say. Basically: if its not a Taylor Sum or some other functional series, the sum of the limits rule does not apply.

Sidharth Ghoshal
  • 18,359