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Question Number 44301 by naka3546 last updated on 26/Sep/18
Provethat1+122+132+142+...=π26
Commented by abdo.msup.com last updated on 26/Sep/18
letconsiderf(x)=∣x∣(2πperiodiceven)letdeveloppfatfourierserief(x)=a02+∑n=1∞ancos(nx)an=2T∫[T]f(x)cos(nx)dx=22π∫−ππ∣x∣cos(nx)dx=2π∫0πxcos(nx)dx⇒π2an=∫0πxcos(nx)dx=[xnsin(nx)]0π−∫0π1nsin(nx)dx=−1n∫0πsin(nx)dx=−1n[−1ncos(nx)]0π=1n2{(−1)n−1}⇒π2an=(−1)n−1n2an=2π(−1)n−1n2⇒π2a0=∫0πxdx=[x22]0π=π22⇒a02=π2∣x∣=π2+2π∑n=1∞(−1)n−1n2cos(nx)=π2−4π∑n=0∞cos(2n+1)x(2n+1)2x=0⇒π2−4π∑n=01(2n+1)2⇒4π∑n=0∞1(2n+1)2=π2⇒∑n=0∞1(2n+1)2=π28but∑n=1∞1n2=∑p=1∞1(2p)2+∑p=0∞1(2p+1)2=14∑n=1∞1n2+π28⇒34∑n=1∞1n2=π28⇒∑n=1∞1n2=4π224=π26.
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