# Show that <mspace width="thinmathspace" /> f ( x ) = <munderover> &#x2211;<!-- ∑

Show that $\phantom{\rule{thinmathspace}{0ex}}f\left(x\right)=\sum _{n=0}^{\mathrm{\infty }}{a}_{n}{x}^{n}, forx\in \left[0,1\right], is of bounded variation$
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Let
$0={x}_{0}<\cdots <{x}_{n}=1,$
then
$\sum _{k=1}^{n}|\phantom{\rule{thinmathspace}{0ex}}f\left({x}_{k}\right)-f\left({x}_{k-1}\right)|=\sum _{k=1}^{n}|\sum _{j=0}^{\mathrm{\infty }}{a}_{j}{x}_{k}^{j}-\sum _{j=0}^{\mathrm{\infty }}{a}_{j}{x}_{k-1}^{j}|\le \sum _{k=1}^{n}\sum _{j=0}^{\mathrm{\infty }}|{a}_{j}|\left({x}_{k}^{j}-{x}_{k-1}^{j}\right)\phantom{\rule{0ex}{0ex}}=\sum _{j=0}^{\mathrm{\infty }}|{a}_{j}|\sum _{k=1}^{n}\left({x}_{k}^{j}-{x}_{k-1}^{j}\right)=\sum _{j=0}^{\mathrm{\infty }}|{a}_{j}|<\mathrm{\infty }.$