function 1/(ln(n+1)) is true

Question

#9 Which of the following statements about convergence of the series (Series 1 is the lower limit ; ∞ is the upper limit ; function 1/(ln(n+1)) is true ????

a)1 is lower limit ; ∞ is the upper limit ; function 1/(ln(n+1) converges by comparison lower limit ; ∞ is the upper limit ; function 1/n

b) 1 is lower limit ; ∞ is the upper limit ; function 1/(ln(n+1) converges by comparison with lower limit ; ∞ is the upper limit ; function 1/n^2

c) 1 is lower limit ; ∞ is the upper limit ; function 1/(ln(n+1) diverges by comparison with lower limit ; ∞ is the upper limit ; function 1/n

d) 1 is lower limit ; ∞ is the upper limit ; function 1/(ln(n+1) diverges by comparison with lower limit ; ∞ is the upper limit ; function 1/n^2

Answer 1

You can eliminate (a) and (d) immediately. Since ∑(1/n) diverges, there's no way that you can use it to show that any series converges by comparison. And since ∑(1/n²) converges, there's no way that you can use it to show that any series diverges by comparison.

Note that ln(n + 1) < n for all n ≥1. That means that 1/ln(n + 1) > 1/n for all n≥1. Therefore since ∑(1/n) diverges, ∑1/ln(n + 1) diverges also.

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