Does the harmonic series ∑ 1/n diverge or converge?

Does the harmonic series ∑ 1/n diverge or converge?

• A. Diverges
• B. Converges
• C. Converges absolutely
• D. Converges conditionally

Answer: (A)

The key idea here is that summing positive terms that decrease as 1/n can still fail to add up to a finite value. The integral test gives a clear check: compare the series to the area under y = 1/x from x = 1 to infinity. The improper integral ∫_1^∞ 1/x dx equals ln x evaluated from 1 to ∞, which diverges to infinity. When the integral diverges for a series with positive terms, the series itself also diverges. Intuitively, the partial sums grow without bound, roughly like ln N, even though the individual terms go to zero. So this harmonic series does not converge in any sense; it diverges.

The key idea here is that summing positive terms that decrease as 1/n can still fail to add up to a finite value. The integral test gives a clear check: compare the series to the area under y = 1/x from x = 1 to infinity. The improper integral ∫_1^∞ 1/x dx equals ln x evaluated from 1 to ∞, which diverges to infinity. When the integral diverges for a series with positive terms, the series itself also diverges. Intuitively, the partial sums grow without bound, roughly like ln N, even though the individual terms go to zero.

So this harmonic series does not converge in any sense; it diverges.