Divide the numerator by the denominator .
Notice that
So as
you have
. Clearly
must converge.
The second sequence requires a bit more work.
The monotone convergence theorem will help here; if we can show that the sequence is monotonic and bounded, then
will converge.
Monotonicity is often easier to establish IMO. You can do so by induction. When
, you have
Assume
, i.e. that
. Then for
, you have
which suggests that for all
, you have
, so the sequence is increasing monotonically.
Next, based on the fact that both
and
, a reasonable guess for an upper bound may be 2. Let's convince ourselves that this is the case first by example, then by proof.
We have
and so on. We're getting an inkling that the explicit closed form for the sequence may be
, but that's not what's asked for here. At any rate, it appears reasonable that the exponent will steadily approach 1. Let's prove this.
Clearly,
. Let's assume this is the case for
, i.e. that
. Now for
, we have
and so by induction, it follows that
for all
.
Therefore the second sequence must also converge (to 2).
The obtuse interior angle in triangle can reach to 179.9999999.... °.
Because once it hits 180°, the angle becomes a straight line and it is not possible in a triangle.
