# Sequences and series: exercise sheet 5

Exercise.

For each of the following, decide (with justification) which of the following series converge.

(a) $∑ n = 1 ∞ n + 1 n 3 - 2$

(b) $∑ n = 1 ∞ n - 1 n 2 - 2$

(c) $∑ n = 1 ∞ n 2 + n - 1 n 3$

(d) $∑ n = 1 ∞ n 2 + n - 1 n 4 - 3 n 2 + 1$

(e) $∑ n = 1 ∞ ( n + 1 ) 2 - n 2 n 3 - n 2 + 17$

(f) $∑ n = 2 ∞ ( n 2 + n - 1 ) 1 2 ( n 5 - 2 ) 1 2$

(g) $∑ n = 2 ∞ ( n 2 + n - 1 ) 1 2 ( n 3 - 2 ) 1 2$

(h) $∑ n = 1 ∞ n 2 n$ (Use the ratio test.)

(i) $∑ n = 1 ∞ n n n$ (You may find it helpful to recall that $( 1 + 1 n ) n → e = 2.71828...$ as $n → ∞$.)

(j) $∑ n = 1 ∞ 2 n + 3 n 4 n + 5 n$

Exercise.

(a) Prove that $∑ n = 1 ∞ n x n$ diverges for all $x > 0$. (Use the null sequence test.)

(b) Prove that $∑ n = 1 ∞ n n 2 + 1 x n$ diverges for all $x > 1$ and converges for $0 < x < 1$. (Use the ratio test.)

(c) Prove that $∑ n = 1 ∞ x n n$ converges for all $x > 0$.

Exercise.

Let $a , b > 0$. Show that

(a) $∑ n = 1 ∞ a + n b + n x n$ converges when $0 < x < 1$ and diverges when $x ⩾ 1$.

(b) $a b x + 2 ( a + 1 ) b 2 x 2 + 3 ( a + 2 ) b 3 x 3 + …$ converges when $0 < x < b$ and diverges when $x ⩾ b$.

Exercise.

Discuss the convergence of

$1 2 x + 1 · 3 2 · 5 x 2 + 1 · 3 · 5 2 · 5 · 8 x 3 + 1 · 3 · 5 · 7 2 · 5 · 8 · 11 x 4 + …$

## Solution to selected parts of these exercises

### Exercise 5.1(a)

The series has positive terms for $n ⩾ 2$ and

$n + 1 n 3 - 2 ⩽ 2 n n 3 / 2 = 4 n 2$

for $n ⩾ 2$, so $∑ n = 2 ∞ n + 1 n 3 - 2$ converges by comparison test with $∑ n = 2 ∞ 1 n 2$ hence $∑ n = 1 ∞ n + 1 n 3 - 2$ also converges. (OR: you could use the limit comparison test here.)

### Exercise 5.1(b)

The series has positive terms for $n ⩾ 2$ and

$n - 1 n 2 - 2 ⩾ n / 2 2 n 2 = 1 4 1 n$

for $n ⩾ 2$, so $∑ n = 2 ∞ n - 1 n 2 - 2$ diverges by comparison test with $∑ n = 2 ∞ 1 n$ hence $∑ n = 1 ∞ n - 1 n 2 - 2$ also diverges. (OR: you could use the limit comparison test here.)

### Exercise 5.3(a)

For $1 > x > 0$, the terms in the series $∑ n = 1 ∞ a + n b + n x n$ are all positive. Let $a n = a + n b + n x n$. We apply the ratio test.

$a n + 1 a n = a + n + 1 b + n + 1 x n + 1 b + n a + n x - n = ( a n + 1 + 1 n ) ( b n + 1 + 1 n ) ( b n + 1 ) ( a n + 1 ) x$

which converges to $x$ by continuity of the arithmetic operations and $1 n → 0$. Therefore by the ratio test, the series converges if $x < 1$ and diverges if $x > 1$. If $x = 1$ the series is $∑ n = 1 ∞ a + n b + n$ and $a + n b + n = a / n + 1 b / n + 1 → 1$ so the series diverges by the null sequence test.