Q.38 Which of the following statements is true for the series given below?                                         sn = 1 + 1/√2 + 1/√3 + 1/√4 + ... + 1/√n sn converges to log(√n) sn converges to √n sn converges to exp(√n) sn diverges

Q.38 Which of the following statements is true for the series given below?

                                        sn = 1 + 1/√2 + 1/√3 + 1/√4 + … + 1/√n

  1. sn converges to log(√n)
  2. sn converges to √n
  3. sn converges to exp(√n)
  4. sn diverges

The partial sum sn = 1 + 1/√2 + 1/√3 + ⋯ + 1/√n represents a p-series with p=1/2 < 1. This series diverges as n approaches infinity.

CORRECT ANSWER

sn diverges. The infinite series k=1 1/√k diverges because it fails the integral test: 1 1/√x dx = limb→∞ 2√b – 2 = ∞.

 Detailed Option Analysis

 Option 1: sn converges to log(√n)

Log(√n) simplifies to (1/2)log n, which grows without bound, but this matches the standard harmonic series Hn ≈ log n + γ, not the p=1/2 case where sn ≈ 2√n.

This option confuses it with the divergent harmonic series Hn.

 Option 2: sn converges to √n

While sn grows like 2√n + C for some constant C (from Euler-Maclaurin summation), √n alone underestimates the growth.

“Converges to” implies a finite limit, which never occurs. The series still diverges.

 Option 3: sn converges to exp(√n)

Exp(√n) grows much faster than any polynomial or root behavior of sn. No series partial sum converges to such a rapidly increasing function.

 Option 4: sn diverges

This holds true. Bounds show 2(√n – 1) < sn < 1 + 2√n, both tending to infinity as n → ∞.

 Mathematical Proof Summary

Integral Test: f(x) = 1/√x is positive, continuous, decreasing for x ≥ 1

Improper Integral:1 x-1/2 dx = [2x1/2]1 = ∞

p-Series Test: p = 1/2 < 1 → diverges

Asymptotic: sn ∼ 2√n + ζ(1/2) + 1/(2√n) + O(1/n3/2)

 

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