Question

Find sum of {(1/3n+1)+(1/3n-1)} from n=1 to infinity.

Answer 1

Answer:

You know the limit (it is indeed log(4/3)), but you probably want to know why it is what it is.

The easiest way is probably to compare the sum to the integral ∫dx/x from 3 to 4, which is of course log(4/3). Cut up the interval [3, 4] into n subintervals; the corresponding right (lower, since the integrand is monotonically decreasing) Riemann sum is 1/n * (1/(3+1/n) + 1/(3+2/n) +1/(3+1/n) +… +1/(3+n/n)) = 1/(3n+1) + 1/(3n+2) + 1/(3n+3) + … + 1/(3n+n). Since the integrand is continuous, the Riemann sum must tend to the integral when n tends to infinity.

Without using fancy language, just think of the integral as the area under the graph y=1/x between the lines x=3 and x=4. Cut it up into n equal vertical slices. The area of the k-th slice, which extends from x=3+(k-1)/n to x=3+k/n and is 1/n wide and whose non-constant height lies between 1/(3+(k-1)/n) and 1/(3+k/n), will be between 1/n * [1/(3+(k-1)/n)] = 1/[3n + (k-1)] and 1/n * 1/(3+k/n) = 1/[3n + k]; the sum of the areas of all n slices will therefore lie between 1/(3n) + 1/(3n+1) + 1/(3n+2) + … + 1/(3n+n-1) and 1/(3n+1) + 1/(3n+2) + 1/(3n+3) + … + 1/(3n+n). The difference of these two sums is 1/(3n)-1/(4n) - 1/(12n), which tends to zero as n tends to infinity; so both those sums are within 1/(12n) of the exact area of our region, which is log(4/3), and must therefore tend to that value as n tends to infinity.

Step-by-step explanation:

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