Question

The value of 1 - 1/2 + 1/3 - 1/4.... infinity
1) ln 2
2) log 2 with base 10
3) e ^2
4) e

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Answer 1

Answer:

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Step-by-step explanation:

1–1/2)=1/2

(1–1/2)(1–1/3)=1/2×2/3=1/3

(1–1/2)(1–1/3)(1–1/4)=1/2×2/3×3/4=1/3×3/4=1/4

(1–1/2)(1–1/3)(1–1/4)(1–1/5)=1/2×2/3×3/4×4/5=1/4×4/5=1/5

As you can see, the denominator of each term cancels out the numerator of the next term, so this series can be generalised to:

(1–1/2)(1–1/3)(1–1/4)…(1–1/n)=1/n

Which can also be written as:

1/2×2/3×3/4×…×((n-1)/n)=1/n

Which we could even be further generalised to:

a/b×b/c×c/d×…×m/n = a/n

Answer 2

Answer:

The correct answer is log 2

Step-by-step explanation:

here is the explanation:

Proof: The infinite series

1–1/2+1/3–1/4+1/5–1/6+……………………………………………………….(1)

is alternating since its terms are alternately positive and negative. Each term is numerically less than the preceding term and the terms decrease indefinitely. Putting it formally, nth term u(n) is monotonic decreasing, i.e. u(n) > u(n+1) for all values of n and u(n) = (-1)^(n-1)/n → 0 as n → ∞ . Hence the series is convergent. Since the series is convergent, it has a finite sum that can be obtained by putting x=1 in the Logarithmic Series (which is known to be convergent in the interval -1< x ≤1)

log (1+x) = x - x² + x³/3 - x^4/4 +……………

and the sum is log 2.

Alternatively, the sum can be determined using the result, as n → ∞, that

1+1/2+1/3+1/4+1/5+….+1/n - log n → γ

where γ is a constant (Euler’s constant). The above result can be rewritten as

1+1/2+1/3+1/4+1/5+….+1/n = log n + γ(n)………………………………………(2)

where γ(n) → γ as n → ∞ . Let s(n) denote the algebraic sum of its first n terms; s the sum of the series. Then

s = lim s(n) as n → ∞ = lim s(2n) as n → ∞ = lim (1–1/2+……-1/2n) as n → ∞

= lim [(1+1/2+1/3+….+1/2n) -2 (1/2+1/4+….+1/2n)] as n → ∞

= lim [(1+1/2+1/3+….+1/2n) - (1+1/2+….+1/n)] as n → ∞ which using relation (2) ,

= lim [log 2n + γ(2n) - log n -γ(n)] as n → ∞

= lim [log 2+ log n +γ(2n)- log n -γ(n)] as n→∞ which,on cancellation of log n term,

= lim [log 2+ + γ(2n) - γ(n)] as n→∞

→ log 2 + γ - γ [since γ(2n) - γ(n) → γ - γ = 0]

Ans= log 2

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