# Is there any way of visually interpreting Weierstrass M test?

Can anyone suggest me some way to visualize Weierstrass M test of uniform convergence of a series of functions?

Statement:

Suppose $${f_n}$$ is a sequence of functions such that $$\exists$$ a sequence $$(M_n)$$ such that $$|f_n(x)|\leq M_n \forall x$$ and $$\forall n \in \Bbb N$$.If $$\sum M_n$$ converges then $$\sum f_n$$ converges uniformly.

I want to visualize graphically what is going on.

• This is not a visual argument, but the following thought process seems fairly clear to me. If $f = \sum_{i=1}^\infty f_i$, then for any $x$ we have $| f(x) - \sum_{i=1}^N f_i(x) | = |\sum_{i=N+1}^\infty f_i(x) | \leq \sum_{i=N+1}^\infty | f_i(x) | \leq \sum_{i=N+1}^\infty M_i$. This gives us a bound on $| f(x) - \sum_{i=1}^N f_i(x) |$ which does not depend on $x$. – littleO Jul 27 '19 at 5:38
• That is nice.But I am still in search of a visual argument. – Kishalay Sarkar Jul 27 '19 at 7:31

• Put $$S_n = 2\sum_{k=1}^n M_k$$, $$S = 2\sum_{k=1}^{\infty} M_k$$, and draw the lines $$y = S_n$$ for each $$n$$. (Think of a rainbow cake of finite thickness $$S$$ but infinitely many layers.)
• Let the $$n$$th layer be the plane between the lines $$y=S_{n-1}$$ and $$y=S_n$$. Put $$Z_n = S_n - M_n$$ for each $$n$$ and draw a dotted line $$y=Z_n$$ for each $$n$$. (The idea is that for each $$n$$ we can shift the $$x$$-axis to the $$n$$th dotted line, and the $$n$$th layer represents ‘$$0 \pm M_n$$’.)
• If for all $$n$$, you can plot $$y=|f_n(x)|+Z_n$$ within the $$n$$th layer then the M test is passed: you can add all of the $$f_n$$ together and the sum will converge.
Uniform convergence is (I think) guaranteed because the layers are the same thickness across the entire domain. For pointwise convergence, the layer thicknesses could vary with $$x$$, and so you would need to accumulate more layers (increase $$n$$) to guarantee that the total thickness is under control.