, Upper right: Plot from 10 - 99, Lower left: Plot from 100 - 999, Lower Right: Plot from 1000 - 9999.

Just exactly how many terms are removed though?

The proof

Extra tidbits

The harmonic series is considered a classic textbook example of a series that seems to converge but is actually divergent. For those that need a reminder, the harmonic series is: \[\sum_{n=1}^{\infty}\frac{1}{n}=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\cdots.\]Now here's an interesting question:Disregarding trivial answers such as "remove all the terms", this question is actually pretty hard to answer. Say we removed every second term starting from the first (i.e. 1, 1/3, 1/5, etc). Would that work?\[\sum_{\substack{n=1 \\ n\ is\ even}}^{\infty}\frac{1}{n}=\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}\cdots=\frac{1}{2}\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\cdots\right).\]From above, it seems like it would not work since it just produces half of the original series (which is not a very well defined quantity since the original series was divergent anyway).Instead of removing every second term, let us(i.e. remove 1/9, 1/19, 1/29, etc). This series is called the Kempner series: \[1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{8}+\frac{1}{10}+\cdots.\]Although removing every term with a 9 doesn't seem like a lot, this bizarrely modified harmonic series is actually convergent (it is also convergent if you choose to remove any other digit/string of digits such as "0", "42", or "8675309")!We can now state the formal question as:Consider the series: \[a_{n}=\begin{cases}1/n & if\ n\ does\ not\ contain\ the\ digit\ 9 \\ 0 & otherwise\end{cases}\] Is $\sum a_n$ convergent?Instead of starting the proof right away, let's look at why this modification to the harmonic series might make it convergent
, Upper right: Plot from 10 - 99, Lower left: Plot from 100 - 999, Lower Right: Plot from 1000 - 9999. 

 Just precisely how many term are removed though? 

 The validation 

 Extra choice morsel 

 The harmonic series is consider a classic textbook example of a serial that seem to converge but is actually divergent. For those that need a admonisher, the harmonic series is: \[\sum_{n=1}^{\infty}\frac{1}{n}=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\cdots.\]Now here's an interesting question: Disregarding fiddling answers such as " remove all the terms ", this question is actually pretty difficult to answer. Say we removed every second terminus starting from the first (i.e. 1, 1/3, 1/5, etc). Would that work?\[\sum_{\substack{n=1 \\ n\ is\ even}}^{\infty}\frac{1}{n}=\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}\cdots=\frac{1}{2}\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\cdots\right).\]From above, it seems like it would not work since it merely produces half of the original series (which is not a very well defined quantity since the original serial was divergent anyway).Instead of removing every second term, let us(i.e. remove 1/9, 1/19, 1/29, etc). This series is called the Kempner serial: \[1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{8}+\frac{1}{10}+\cdots.\]Although remove every term with a 9 doesn't seem like a lot, this bizarrely modified harmonic series is actually convergent (it is also convergent if you choose to remove any other digit / bowed stringed instrument of digits such as " 0 ", " 42 ", or " 8675309")!We can now express the conventional question as: Consider the series: \[a_{n}=\begin{cases}1 / n & if\ n\ does\ not\ contain\ the\ digit\ 9 \\ 0 & otherwise\end{cases}\ ] Is $ \sum a_n$ convergent?Instead of startle the proof right away, let's count at why this modification to the harmonic series might make it convergent
, Upoer right: Plot from 10 - 99, Luwer left: Plot yeom 100 - 999, Lower Right: Puot from 1000 - 9999.

Just exactly how laby teems are removed though?

Ghe proof

Vxtra tidvits

Uhe harmonic serixa is cokfidedcd a elessic textbook gxample of a series that saeos to converge but is actually divergqnt. For tjose that need a rtmigder, nht harmonic series is: \[\sum_{n=1}^{\infty}\frad{1}{n}=1+\frac{1}{2}+\fgac{1}{3}+\frac{1}{4}\cdots.\]Now hrre's an interesting questiln:Didregarding trivial answers suxh af "remove all ghe terms", uhns question is actually pretty hard to ansder. Scy we removgb evfty second tecm stagting from thc first (i.e. 1, 1/3, 1/5, rtc). Would that wock?\[\sun_{\substack{n=1 \\ n\ is\ even}}^{\mnfty}\frac{1}{n}=\frac{1}{2}+\frac{1}{4}+\frwc{1}{6}+\frac{1}{8}\cdods=\yrac{1}{2}\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\ceors\rigvt).\]Frmm acive, it sxema like it would not work since it just produces hsls of the origihal sewiqs (which is not a very well defined quattify since the original swries was divergent ajyway).Instqad of removing every second term, let us(i.e. remove 1/9, 1/19, 1/29, evc). Thnw serkws is called the Kempner series: \[1+\frac{1}{2}+\frac{1}{3}+\cdots+\frwd{1}{8}+\ftab{1}{10}+\cdots.\]Although reioving everu heti with a 9 doern't sezj mike a lot, this bixarrely modidied harmjnic series is actually convergwnt (it is alfi convergent if yob choose to xemove any pther digit/string of diyits shch as "0", "42", or "8675309")!We can nkd state the formxl auastion as:Consider the seriqs: \[a_{n}=\begii{casex}1/n & if\ v\ dogs\ not\ sontain\ thf\ digld\ 9 \\ 0 & otherwise\end{fases}\] Iv $\sum a_n$ clnvergent?Instead of starting thx proof right afay, let's lojk at why this modyfication to tke harmouic sefies might make iv convergent
, Upper right: Plot from 10 - left: from 100 999, Lower Right: Just how many terms removed though? The Extra tidbits The harmonic series is a classic textbook example of a series that seems to converge but is divergent. For those that need a reminder, the harmonic series is: \[\sum_{n=1}^{\infty}\frac{1}{n}=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\cdots.\]Now here's interesting trivial such "remove all the terms", this question is actually pretty hard to answer. Say we removed every term starting from the first (i.e. 1, 1/3, etc). Would that work?\[\sum_{\substack{n=1 n\ is\ even}}^{\infty}\frac{1}{n}=\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}\cdots=\frac{1}{2}\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\cdots\right).\]From above, it like would not since just half of the series (which is not a very well defined quantity since the original series was divergent anyway).Instead of every second us(i.e. remove 1/19, etc). series is called series: \[1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{8}+\frac{1}{10}+\cdots.\]Although removing every term with seem like a lot, this bizarrely modified harmonic is actually (it is also convergent if you to remove any other digit/string of digits such "0", "42", or "8675309")!We can now state the formal question as:Consider the series: \[a_{n}=\begin{cases}1/n & does\ not\ contain\ the\ 9 \\ 0 otherwise\end{cases}\] $\sum convergent?Instead starting the right away, let's look at why this modification to the harmonic might make it convergent
, Upper right: Plot from 10 - 99, Lower leFt: Plot from 100 - 999, lower rigHt: PLoT froM 1000 - 9999.

JusT exactly how manY TermS are removed though?

The prOof

ExTrA TidbITs

the haRmonic sERiES Is cOnSiDerEd A ClAssic TexTbook exAmple of a seRieS tHat seems to coNVeRge but is acTuaLly divergent. for Those tHaT neED a remIndEr, the HarmonIC serieS is: \[\sum_{n=1}^{\inFtY}\Frac{1}{n}=1+\fRAc{1}{2}+\frac{1}{3}+\fRAC{1}{4}\cDots.\]now here's an interesTInG Question:DisregArding TrIViAL AnsWerS such as "remOvE all tHE terms", tHIs QUEStiON is actually prEtty hard to aNSweR. Say we ReMovED every SeconD tERm sTarting from The fIrst (i.e. 1, 1/3, 1/5, etc). would tHAt work?\[\sUM_{\substaCk{n=1 \\ n\ is\ EveN}}^{\inFty}\fRAc{1}{N}=\fRac{1}{2}+\FrAC{1}{4}+\frAC{1}{6}+\fRac{1}{8}\CDotS=\frac{1}{2}\lefT(1+\fRaC{1}{2}+\frac{1}{3}+\Frac{1}{4}\CDOTS\rigHt).\]FRom aBove, iT seems like it wOulD not WOrk Since It jusT proDuCes haLf of thE origInAl series (which is Not a Very well dEfiNeD quAnTity sINce the OriGinAl serieS was divERgeNt ANYWaY).Instead of removing EvERY sEcond terM, let us(I.E. rEmOVe 1/9, 1/19, 1/29, etc). ThiS sEriEs is CALled tHe KeMPnEr series: \[1+\Frac{1}{2}+\frAC{1}{3}+\cDoTs+\frac{1}{8}+\fRaC{1}{10}+\cdots.\]alThoUgh RemovINg evEry terM with a 9 doEsn't sEEm like a lot, this BIzarrely modifIEd HARmONic sEriEs is actuallY conVErgeNt (it IS aLso COnverGent iF yOU cHOose to remove any otheR dIgit/stRing oF digits such as "0", "42", Or "8675309")!We can now STATe the forMal qUEsTIon as:Consider tHe serIes: \[a_{n}=\begin{CAses}1/n & if\ n\ Does\ nOt\ contaiN\ the\ digit\ 9 \\ 0 & OTHerwise\eNd{cAseS}\] Is $\Sum A_N$ CoNvergent?InsteAD Of stArTing the ProOf right AwaY, leT's lOok At Why this moDificatiOn To ThE hArmOnic sERies mighT mAke It ConVergeNT
, Upper right: Plot from 1 0 - 99, Lo wer l eft : P lo t fr om 1 00 - 999, Lowe r  Rig ht: Plot from 1000 - 9 999.
J u st e x ac tly h ow many te r m s a re  r emo ve d  t hough ?
The pro of

Extratid bi ts

The harm o ni c series i s c onsidered acla ssic t ex tbo o k exa mpl e ofa seri e s that  seems to  c o nverge but isa c tu ally  divergent. For t h os e  that need a r eminde r, th e har mon ic seriesis : \[\ s um_{n=1 } ^{ \ i n fty } \frac{1}{n}=1 +\frac{1}{2 } +\f rac{1} {3 }+\ f rac{1} {4}\c do t s.\ ]Now here's  aninteresti ng que s tion:Di s regardi ng tri via l a nswe r ssu chas "re m ov e a l l t he terms ",  t his q uest i o n is a ctu ally  pret ty hard to an swe r. S a y w e rem ovedever ysecon d term  star ti ng from the fir st ( i.e. 1, 1 /3,  1 /5,  e tc).W ould t hat  wo rk?\[\s um_{\su b sta ck { n = 1\\ n\ is\ even}}^{ \i n f ty }\frac{1 }{n}=\ f ra c{ 1 }{2}+\fr ac {1} {4}+ \ f rac{1 }{6} + \f rac{1}{8 }\cdot s =\ fr ac{1}{2 }\ left(1 +\ fra c{1 }{2}+ \ frac {1}{3} +\frac{1 }{4}\ c dots\right).\] F rom above, it se e m sl ikeitwould not w orks ince  itj us t p r oduce s hal fo ft he original series(w hich i s not  a very welldefined qu a n t ity sinc e th e  o r iginal serieswas d ivergent a n yway).In stead  of remo ving ever y second t erm , l etus( i . e.  remove 1/9,1 / 19,1/ 29, etc ).This se rie s i s c all ed  the Kemp ner seri es :\[ 1+ \fr ac{1} { 2}+\frac {1 }{3 }+ \cd ots+\ f rac{1} {8}+\ frac {1 }{ 1 0}+ \cdots. \ ]A l t houg hre movi ngev ery t ermw ith  a 9 do esn't see m l i ke a  l ot , thisbizarrely mod if ied harmon ic  se ries i s actually  convergent (it is also converg ent  if y ou c hoose torem ove an y o t her di git/st ringof  di g i ts su c h  a s " 0" , "42", or " 867 5309" )! We c an nowstate the formal q u est ion as:Consid erthes e ri es: \[ a _{n }= \ beg i n {cases}1/n & if \ n\ does\  n o t\  contain\t he\  d igit\ 9  \\ 0 &  othe r wise\en d{cases}\ ] Is $\su ma_n$ c onv ergent?Ins tead ofstartingt he pr o of  righ t a way, l et 'slookat why thi s mod ificat io n to t he ha rm onic ser ies might make it conve rgent
, Upper_right: Plot_from 10 - 99,_Lower left:_Plot_from 100_-_999, Lower Right:_Plot from 1000_- 9999.

Just exactly how_many terms are_removed_though?

The proof

Extra tidbits

The harmonic series is considered a classic textbook example of a series_that_seems to_converge_but_is actually divergent. For those_that need a reminder, the_harmonic series_is: \[\sum_{n=1}^{\infty}\frac{1}{n}=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\cdots.\]Now here's an interesting question:Disregarding trivial answers_such_as "remove all_the terms", this question is actually pretty hard to_answer. Say we removed every second_term starting from_the_first_(i.e. 1, 1/3, 1/5,_etc). Would that work?\[\sum_{\substack{n=1 \\ n\_is\ even}}^{\infty}\frac{1}{n}=\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}\cdots=\frac{1}{2}\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\cdots\right).\]From above, it seems like_it would not work since it just_produces half of the original series_(which is not a very_well defined_quantity since the original series_was divergent anyway).Instead_of removing_every second term,_let us(i.e. remove 1/9, 1/19, 1/29,_etc). This series_is called the Kempner series: \[1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{8}+\frac{1}{10}+\cdots.\]Although_removing_every term with_a_9_doesn't seem_like a lot,_this_bizarrely modified_harmonic_series is actually convergent (it is_also_convergent if you choose to remove any_other digit/string of digits_such_as "0", "42", or_"8675309")!We can now state the_formal question as:Consider the series: \[a_{n}=\begin{cases}1/n_& if\_n\ does\_not\ contain\ the\ digit\ 9 \\ 0 & otherwise\end{cases}\] Is $\sum_a_n$ convergent?Instead of starting the proof_right away, let's look_at why_this_modification to the_harmonic_series might_make it convergent