Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2009 / Paper 1 / 20 / a

Table of contents
- i a
  - Solution i
- ii a
  - Solution ii
- iii a
  - Solution iii

i

Solution

The summand is

n / n + 1 = n / (n + 1) \to 1

, rather than zero. The necessary term test for divergence fails. Therefore the positive-term series diverges to $+ \infty$ .

ii

Solution

For

a_{n} = n^{2} / n!

, the ratio test gives

\frac{a _{n + 1}}{a _{n}} = \frac{( n + 1 ) ^{2}}{n ^{2} ( n + 1 )} = \frac{n + 1}{n ^{2}} ⟶ 0 < 1.

Thus the series converges absolutely. As a check,

n^{2} = n (n - 1) + n

and the exponential power series give the sum

e + e = 2 e

.

iii

Solution

For

a_{n} = n^{n} / n!

, successive terms satisfy

\frac{a _{n + 1}}{a _{n}} = (1 + \frac{1}{n})^{n} ⟶ e > 1.

They eventually grow by a factor strictly greater than one and cannot tend to zero. Hence the series diverges to $+ \infty$ , by the ratio test or the term test for divergence.

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