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{SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT -1 216 "Problem: We know sum(1/n^
2,n=1..infinity) is convergent and converges to Pi^2/6 (by using Maple
). Find the smallest integer N so that the difference between its part
ial sum (up to N) and Pi^2/6  is less than 10^(-3)." }}{PARA 11 "" 1 "
" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "L:=sum(1/
n^2,n=1..infinity);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"LG,$*&\"\"'
!\"\"%#PiG\"\"#\"\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "f:
=N->sum(1/n^2,n=1..N);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"fGf*6#%
\"NG6\"6$%)operatorG%&arrowGF(-%$sumG6$*&\"\"\"F0*$)%\"nG\"\"#F0!\"\"/
F3;F09$F(F(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "evalf(f(10
0));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"++R)\\j\"!\"*" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "fsolve(abs(L-f(N))=0.001,N);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+n\"**\\***!\"(" }}}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 43 "evalf(abs(f(999)-L));evalf(abs(f(1000)-L));
" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"(,0+\"!\"*" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#$\"+n;+&***!#8" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 62 "
So the smallest integer such that abs(f(N)-L)<10^(-3) is 1000." }}}}
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