A ratio of two series blows up at certain $\epsilon$ values. Why?

Question

I am trying to understand the behaviour of this function:

$$f(x)=\frac{\frac{1}{2}-\frac{4}{1+\epsilon hx}\sum_{n=0}^{\infty}\frac{(-1)^n}{\lambda_n^4}tanh\bigg(\frac{\lambda_n(1+\epsilon hx)}{2}\bigg)}{\frac{1}{3}-\frac{4}{1+\epsilon hx}\sum_{n=0}^{\infty}\frac{1}{\lambda_n^5}tanh\bigg(\frac{\lambda_n (1+\epsilon hx)}{2}\bigg)}$$

Where $\lambda_n=(n+\frac{1}{2})\pi$, and h is a constant.

For $\epsilon=0$, I get a constant horizontal plot, which makes sense.

The more I increase the value of $\epsilon$, the more the graph starts to descend, for example in this case of $\epsilon=0.00001$:

Graph of $f(x)$ for $\epsilon=0.00001$

But at a certain point, the graph starts to fluctuate at a certain $x$ position and blows up. For example, for $\epsilon=0.001$:

Graph of $f(x)$ for $\epsilon=0.0001$

It's not an asymptote, the peak of the blow up is at around 6,000, and the more I decrease the $\epsilon$ value, the lower the peak is at. My question is this: Why does this behaviour occur?

Thanks in advance!

Answer 1

It is difficult to say since we do not have any idea about the values of $h$ and $x$ for which you have the problem.

So, let $\delta=\epsilon h x$ and consider $F(\delta)=\frac AB$ with $$A=\frac{1}{2}-\frac{4}{1+\delta}\sum_{n=0}^{\infty}\frac{(-1)^n}{\lambda_n^4}\tanh\bigg(\frac{\lambda_n(1+\delta)}{2}\bigg)$$ $$B=\frac{1}{3}-\frac{4}{1+\delta}\sum_{n=0}^{\infty}\frac{1}{\lambda_n^5}\tanh\bigg(\frac{\lambda_n (1+\delta)}{2}\bigg)$$ and for conveniency, let me use $\delta=10^k$.

I give below a table of my results $$\left( \begin{array}{cccc} k & A & B & \frac AB \\ 6 & 0.49999935 & 0.33333291 & 1.49999994 \\ 5 & 0.49999350 & 0.33291359 & 1.50187172 \\ 4 & 0.49993503 & 0.33329132 & 1.49999415 \\ 3 & 0.49935089 & 0.33291359 & 1.49994145 \\ 2 & 0.49356674 & 0.32917327 & 1.49941316 \\ 1 & 0.44093099 & 0.29513643 & 1.49399036 \\ 0 & 0.20233616 & 0.14057702 & 1.43932605 \\ -1 & 0.08944269 & 0.06611009 & 1.35293550 \\ -2 & 0.07755340 & 0.05805909 & 1.33576667 \\ -3 & 0.07637861 & 0.05725922 & 1.33390940 \\ -4 & 0.07626130 & 0.05717930 & 1.33372220 \\ -5 & 0.07624957 & 0.05717131 & 1.33370346 \\ -6 & 0.07624840 & 0.05717051 & 1.33370159 \\ -7 & 0.07624828 & 0.05717043 & 1.33370140 \\ -8 & 0.07624827 & 0.05717042 & 1.33370138 \\ -9 & 0.07624827 & 0.05717042 & 1.33370138 \\ -10 & 0.07624827 & 0.05717042 & 1.33370138 \end{array} \right)$$

Answer 2

It seems the denominator is the problematic element here.

For the case of $\epsilon=0.00001$ and $h=10^6$, the denominator never approaches $0$:

Graph of the denominator for $\epsilon=0.00001$

But as I increase the $\epsilon$ value to $\epsilon=0.0001$, it hits zero at $x=0.11$, which is exactly where the blow up occurs.

Graph of the denominator for $\epsilon=0.0001$

But now the question is, if the denominator does reach $0$ at $x=0.11$, why does it blow up to a finite number? Shouldn't there be an aymptote at that location for $f(x)$?