Question

Prove that , 1 + 2 + 3 + 4 + _____+ ∞ = -1/12

( Ramanujan's Summation to Infinity Theory . ) ​

Answer 1

TO PROVE :–

$\\ \bf \implies1 + 2 + 3 + 4 +. . . . . . + \infty = - \dfrac{1}{12}$

SOLUTION :–

• Let a series –

$\\ \bf \implies S_1 =1 - 1 + 1 - 1 + . . . . . .$

• If number of terms are Even –

$\\ \bf \implies S_1 =0 \: \: \: \: \: \: \: \: \: \: \: \: \: - - - eq.(1)$

• If number of terms are odd –

$\\ \bf \implies S_1 =1 \: \: \: \: \: \: \: \: \: \: \: \: \: - - - eq.(2)$

• Now add eq.(1) & eq.(2) –

$\\ \bf \implies S_1 +S_1=0 + 1$

$\\ \bf \implies 2S_1=1$

$\\ \large \bf\implies {\boxed{ \bf S_1= \dfrac{1}{2}}} \: \: \: \: \: \: \: \: \: \: \: \: \: \: - - - eq.(3)$

• Now let's take another series –

$\\ \bf \implies S_2=1 -2+3-4 + . . . . . .$

• Add S₂ in S₂ after shifting one term –

$\green{ \implies} \sf{\red{S_2 = 1 - 2 + 3 + 4 + - - - - }} \\ \qquad \: \: \underline{\sf{\red{S_2 = \quad \quad 1 - 2 + 3 - - - }}} \\ \qquad \: \: \underline{ \sf{\red{ S_2 = 1 - 1 + 1 - 1 + - - - - }}}$

$\\ \bf \implies 2S_2=1 -1+1-1 + . . . . . .$

$\\ \bf \implies 2S_2=S_1$

• By using eq.(3) –

$\\ \bf \implies 2S_2= \dfrac{1}{2}$

$\\ \large \bf\implies { \boxed{ \bf S_2= \dfrac{1}{4} }} \: \: \: \: \: \: \: \: \: \: \: - - - - eq.(4)$

• Now let's take required equation –

$\\ \bf \implies S_3=1 + 2 +3 + 4 + . . . . . .$

• Subtract S₃ from S₂ –

$\green{ \implies} \sf{\red{S_3 = 1 + 2 + 3 + 4 + - - - - }} \\ \qquad \: \: \sf{\red{S_2 = 1 -2 + 3 - 4 + - - - }} \\ \: \underline{ \sf{ \red{ \qquad\: - \quad \: \: - \: \: + \: \: - \: \: + \: \: - \: \: \: \: \: \: \: \: \: \: \: \: \: }}}\\ \: \: \underline{ \sf{\red{ S_3 - S_2 = 4 + 8 + 12 + - - - - }}}$

$\\ \bf \implies S_3 -S_2=4 + 8 + 12+ . . . . . .$

$\\ \bf \implies S_3 -S_2=4(1 + 2 + 3 + 4 + . . . . . . )$

$\\ \bf \implies S_3 -S_2=4(S_3)$

$\\ \bf \implies -S_2=4(S_3) -S_3$

• Using eq.(4) –

$\\ \bf \implies 3(S_3) = - \dfrac{1}{4}$

$\\ \bf \implies S_3 = - \dfrac{1}{12}$

$\\ \large \implies{\boxed{\bf 1 + 2 + 3 + 4 + . . . . . . = - \dfrac{1}{12}}}$

$\\ \large \implies{\boxed{\bf Hence \: \: Proved}}$