# Prove the following series is true?

As n approaches infinity: 1/2+1/4+1/8+1/16+⋯+1/2^n =1

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• ½ = 1 - ½ {as ½ + ½ = 1}

½ + ¼ = 1 - ¼ {as ½ + ¼ + ¼ = 1}

½ + ¼ + ⅛ = 1 - ⅛ {etc.}

½ + ¼ + ⅛ + ¹∕₁₆ = 1 - ¹∕₁₆

½ + ¼ + ⅛ + ¹∕₁₆ + ¹∕₃₂ = 1 - ¹∕₃₂

½ + ¼ + ⅛ + ¹∕₁₆ + ¹∕₃₂ + … + ½ⁿ = 1 - ½ⁿ

as n→∞, ½ⁿ→0, so we can discard the diminishing final term, yielding:[k = 1 to ∞]∑(½ᵏ) = 1

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• S = 1/2 + 1/4 + 1/8 + 1/10 + ………. + 1/2n = 1

S = 1/2 + 1/(2)² + 1/(2)³ + ………… + 1/2.nΣ(S+1) = 1 + 1/2 + 1/(2)² + 1/(2)³ + 1/(2)⁴ + ……….S + 1 = 1 + 1/2 + 1/2² + 1/2³ + ……………..S + 1 = 1/2n.

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• :-

S oo = a / ( 1 - r )

S oo = (1/2) / ( 1 - 1/2 ) = 1

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• 1/2 + 1/4 + 1/8 + 1/16 +⋯+1/2^n = 1

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• Let's say that this sums to S

1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + .... = S

S = 1/2 + (1/4 + 1/8 + 1/16 + 1/32 + 1/64 + ....)

S = 1/2 + 1/2 * (1/2 + 1/4 + 1/8 + 1/16 + 1/32 + ....))

S = 1/2 + (1/2) * S

S - (1/2) * S = 1/2

S * (1 - 1/2) = 1/2

S * (1/2) = 1/2

S = 1

S = 1/2 + 1/4 + 1/8 + 1/16 + ....

S = 1

Therefore

1 = 1/2 + 1/4 + 1/8 + 1/16 + ....

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• sure : 1/4 +1/8 +1/16+1/32...+1/2^n = 1/2

1/2+1/2 = 1.00

• His solution is wrong since it assumes the result. U can always prove an assertion if you assume its true. You should have noticed this error. Look at my solution say; nowhere did I assume that the sum equals 1, yet I arrive at that conclusion.

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• S = 1/2+1/4+1/8+1/16+⋯+1/2^n

2S = 2(1/2+1/4+1/8+1/16+⋯+1/2^n)

2S=1 + (1/2  + 1/4 + ... 1/2^(n-1))

2S=1 + (1/2 + 1/4 + ... 1/2^(n-1) + 1/2ⁿ) - 1/2ⁿ

2S= 1 + S - 1/2ⁿ

Subtracting S each side gives

S = 1 - 1/2ⁿ.

If n is very large, what is S?

Done!

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• S(n) = ½ + ¼ + 1/8 +.....+ 1/2ⁿ

(1/2)S(n) = ¼ + 1/8 +.....+ 1/2ⁿ⁺¹

S(n) - (1/2)S(n) = (1/2)S(n) = ½ - 1/2ⁿ⁺¹

S(n) = 1 - 1/2ⁿ

As n→∞, 1/2ⁿ⁺¹→0 & S(n) →1.

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