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Weierstrass $M$ test states that , If {${f_n}$} is a sequence of functions on $D$ such that $|f_n(x)|<M_n$ for every $x$ in $D$ for some $M_n$.If $M_1+M_2+.....$converges then $f_1+f_2+....$converges uniformly. How can I interpret what it means actually i.e. I want to visualize this theorem.Is there any way to visualize this theorem?

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    $\begingroup$ Think of it as bounding above a sequence of functions by a summable sequence. $\endgroup$ – fGDu94 Jul 23 '19 at 3:05
  • $\begingroup$ what do you mean,please state more precisely. $\endgroup$ – user685348 Jul 23 '19 at 5:15
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Here’s one visualisation (See the statement of Weierstrass M test):

  • 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.

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    $\begingroup$ This question is answered by someone else in my profile so I thought that sharing it will help. $\endgroup$ – Kishalay Sarkar Jul 28 '19 at 1:16

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