Porisma

Worked solution

Separate monotonic behaviour from convergence

Describe the behaviour of the following sequences for n∈Z+n\in\mathbb Z^+: (a) un=2n+1u_n=2n+1; (b) un=(−1/2)nu_n=(-1/2)^n; (c) un=(−2)nu_n=(-2)^n; (d) un=enu_n=e^n; (e) un=2+e−nu_n=2+e^{-n}.

CJC · Sequences and Series · Example 1

(a) Linear growth

Answer

Increasing and divergent.

Justify the printed behaviour

un+1−un=2>0,2n+1→+∞u_{n+1}-u_n=2>0,\qquad 2n+1\to+\infty

The source tables start at 1 and 100. The Teacher emphasises that 100 is an arbitrary sample index, not infinity.

How did this part go?

(b) Shrinking alternation

Answer

Neither increasing nor decreasing; converges to 0.

Justify the printed behaviour

∣un∣=2−n→0,−2−n≤un≤2−n|u_n|=2^{-n}\to0,\qquad -2^{-n}\le u_n\le2^{-n}

Odd terms are negative and even terms positive. Alternation does not prevent convergence when the magnitudes shrink to zero.

How did this part go?

(c) Growing alternation

Answer

Neither increasing nor decreasing; divergent.

Justify the printed behaviour

u2k=22k→+∞,u2k−1=−22k−1→−∞u_{2k}=2^{2k}\to+\infty,\qquad u_{2k-1}=-2^{2k-1}\to-\infty

The two parity subsequences cannot approach one finite limit. The whole sequence does not tend to positive infinity either.

How did this part go?

(d) Exponential growth

Answer

Increasing and divergent.

Justify the printed behaviour

un+1un=e>1,en→+∞\frac{u_{n+1}}{u_n}=e>1,\qquad e^n\to+\infty

All terms are positive; the ratio shows strict increase. The finite calculator table is an illustration of the growth.

How did this part go?

(e) Exponential decay towards two

Answer

Decreasing; converges to 2.

Justify the printed behaviour

un+1−un=e−n(e−1−1)<0,e−n→0,un→2u_{n+1}-u_n=e^{-n}(e^{-1}-1)<0,\qquad e^{-n}\to0,\qquad u_n\to2

The Teacher warns about calculator limitations: the displayed table rounds later terms to 2. Exact terms remain strictly above 2 for every finite positive index.

How did this part go?

Common mistakes

Increasing does not imply convergence. Alternating does not imply divergence. A calculator display equal to 2 may be a rounded value, not an exact term.

Solution notes

The source uses early and late calculator tables to illustrate these behaviours. The algebra here explains why they continue beyond the displayed terms. A finite sample does not prove an infinite limit.