Question

How do you test for convergence $\dfrac{{\sin \left( {2n} \right)}}{{1 + {2^n}}}$ from $n = 1$ to infinity?

Answer 1

Hint: In order to test for the convergence for the function $\dfrac{{\sin \left( {2n} \right)}}{{1 + {2^n}}}$ given, first look for the range of the denominator and numerator, check whether they converge or diverge. Then check the function collectively through which test it converges or diverges.

Complete step by step solution:
We are given with the function $\dfrac{{\sin \left( {2n} \right)}}{{1 + {2^n}}}$ from $n = 1$ to infinity that can be written as:
$\sum\limits_{n = 1}^\infty {\dfrac{{\sin \left( {2n} \right)}}{{1 + {2^n}}}} $
Let’s check for the test for convergence(whether it converges or diverges);
Let’s see the numerator which is $\sin \left( {2n} \right)$ and we know that the range of sine function is from $ - 1$ to $1$.
So, it becomes $ - 1 \leqslant \sin x \leqslant 1$ and for our numerator it becomes $ - 1 \leqslant \sin \left( {2n} \right) \leqslant 1$.
Now, for the denominator we can see that for ${2^n}$ geometric series is followed.
As we know about the range, So collectively writing the absolute value of the function we get:
$|\dfrac{{\sin \left( {2n} \right)}}{{1 + {2^n}}}| \leqslant \dfrac{1}{{1 + {2^n}}}$
And we know that $\dfrac{1}{{1 + {2^n}}} \leqslant \dfrac{1}{{{2^n}}}$ because as the denominator increases its value decreases.
So, we can write it as:
\[|\dfrac{{\sin \left( {2n} \right)}}{{1 + {2^n}}}| \leqslant \dfrac{1}{{1 + {2^n}}} \leqslant \dfrac{1}{{{2^n}}}\]
And,
\[\sum\limits_{n = 1}^\infty {\dfrac{1}{{{2^n}}}} \] converges by the Geometric series test and since, \[r = \dfrac{1}{2} < 1\], where \[r\] is the common ratio as we write \[\dfrac{1}{{{2^n}}} = \dfrac{{{1^n}}}{{{2^n}}} = {\left( {\dfrac{1}{2}} \right)^n} = a{r^n}\].
Now, for the combined value we get $\sum\limits_{n = 1}^\infty {\dfrac{{\sin \left( {2n} \right)}}{{1 + {2^n}}}} $ converges by Direct Convergence Test.
As the larger series converges, the smaller series also converges.
Thus, over all we test for convergence $\dfrac{{\sin \left( {2n} \right)}}{{1 + {2^n}}}$ from $n = 1$ to infinity with D.C.T of Direct Convergence Test and it converges.

Note:
1) If Every preceding term is less than the succeeding term in any convergent series, then it must converge automatically. This is known as D.C.T or Direct Convergence Test.
2) It’s very important to check for the range of the functions given before checking its convergence or divergence.