Week 4 Discussion Notes
Table of Contents
Infinite Series
Definition
The infinite sum of a sequence ( a n ) n \p{a_n}_n ( a n ) n
∑ n = 1 ∞ a n = lim N → ∞ ∑ n = 1 N a n . \sum_{n=1}^\infty a_n = \lim_{N\to\infty} \sum_{n=1}^N a_n. n = 1 ∑ ∞ a n = N → ∞ lim n = 1 ∑ N a n .
For most of this class, we'll be concerned with whether an infinite series converges or diverges, i.e., whether the sequence of partial sums ( ∑ n = 1 N a n ) N \p{\sum_{n=1}^N a_n}_N ( ∑ n = 1 N a n ) N can calculate the actual sum, though:
Geometric Series
Here's one formula for the geometric series:
Theorem
If ∣ r ∣ < 1 \abs{r} < 1 ∣ r ∣ < 1 ∑ n = 0 ∞ r n \sum_{n=0}^\infty r^n ∑ n = 0 ∞ r n
∑ n = 0 ∞ r n = 1 + r + r 2 + ⋯ = 1 1 − r . \sum_{n=0}^\infty r^n
= 1 + r + r^2 + \cdots
= \frac{1}{1 - r}. n = 0 ∑ ∞ r n = 1 + r + r 2 + ⋯ = 1 − r 1 .
There are more formulas that are more general, but in my opinion this is the easiest to remember, and if you remember this one, you can apply it to any geometric series:
Example 1.
Calculate ∑ n = 3 ∞ 5 ⋅ 2 3 n + 2 3 2 n + 3 \displaystyle \sum_{n=3}^\infty 5 \cdot \frac{2^{3n+2}}{3^{2n+3}} n = 3 ∑ ∞ 5 ⋅ 3 2 n + 3 2 3 n + 2
Solution.
To use the formula, we need to rewrite the sum into a form that lets us apply it easily.
Using the basic properties of exponentials, we get
2 3 n + 2 = ( 2 3 ) n ⋅ 2 2 = 8 n ⋅ 2 2 3 2 n + 3 = ( 3 2 ) n ⋅ 3 3 = 9 n ⋅ 3 3 , \begin{aligned}
2^{3n+2}
&= \p{2^3}^n \cdot 2^2
= 8^n \cdot 2^2 \\
3^{2n+3}
&= \p{3^2}^n \cdot 3^3
= 9^n \cdot 3^3,
\end{aligned} 2 3 n + 2 3 2 n + 3 = ( 2 3 ) n ⋅ 2 2 = 8 n ⋅ 2 2 = ( 3 2 ) n ⋅ 3 3 = 9 n ⋅ 3 3 , so the sum becomes
∑ n = 3 ∞ 5 ⋅ 2 3 n + 2 3 2 n + 3 = ∑ n = 3 ∞ 5 ⋅ 8 n ⋅ 2 2 9 n ⋅ 3 3 = 5 ⋅ 2 2 3 3 ∑ n = 3 ∞ ( 8 9 ) n = 5 ⋅ 2 2 3 3 [ ( 8 9 ) 3 + ( 8 9 ) 4 + ( 8 9 ) 5 + ⋯ ] = 5 ⋅ 2 2 3 3 ( 8 9 ) 3 [ 1 + 8 9 + ( 8 9 ) 2 + ⋯ ] = 5 ⋅ 2 2 3 3 ( 2 3 3 2 ) 3 1 1 − 8 9 = 5 ⋅ 2 2 3 3 2 9 3 6 1 1 9 = 5 ⋅ 2 11 3 9 ⋅ 3 2 = 5 ⋅ 2 11 3 7 . \begin{aligned}
\sum_{n=3}^\infty 5 \cdot \frac{2^{3n+2}}{3^{2n+3}}
&= \sum_{n=3}^\infty 5 \cdot \frac{8^n \cdot 2^2}{9^n \cdot 3^3} \\
&= \frac{5 \cdot 2^2}{3^3} \sum_{n=3}^\infty \p{\frac{8}{9}}^n \\
&= \frac{5 \cdot 2^2}{3^3} \br{\p{\frac{8}{9}}^3 + \p{\frac{8}{9}}^4 + \p{\frac{8}{9}}^5 + \cdots} \\
&= \frac{5 \cdot 2^2}{3^3} \p{\frac{8}{9}}^3 \br{1 + \frac{8}{9} + \p{\frac{8}{9}}^2 + \cdots} \\
&= \frac{5 \cdot 2^2}{3^3} \p{\frac{2^3}{3^2}}^3 \frac{1}{1 - \frac{8}{9}} \\
&= \frac{5 \cdot 2^2}{3^3} \frac{2^9}{3^6} \frac{1}{\frac{1}{9}} \\
&= \frac{5 \cdot 2^{11}}{3^9} \cdot 3^2 \\
&= \boxed{\frac{5 \cdot 2^{11}}{3^7}}.
\end{aligned} n = 3 ∑ ∞ 5 ⋅ 3 2 n + 3 2 3 n + 2 = n = 3 ∑ ∞ 5 ⋅ 9 n ⋅ 3 3 8 n ⋅ 2 2 = 3 3 5 ⋅ 2 2 n = 3 ∑ ∞ ( 9 8 ) n = 3 3 5 ⋅ 2 2 [ ( 9 8 ) 3 + ( 9 8 ) 4 + ( 9 8 ) 5 + ⋯ ] = 3 3 5 ⋅ 2 2 ( 9 8 ) 3 [ 1 + 9 8 + ( 9 8 ) 2 + ⋯ ] = 3 3 5 ⋅ 2 2 ( 3 2 2 3 ) 3 1 − 9 8 1 = 3 3 5 ⋅ 2 2 3 6 2 9 9 1 1 = 3 9 5 ⋅ 2 11 ⋅ 3 2 = 3 7 5 ⋅ 2 11 .
Telescoping Series
The other main type of series where you can calculate sums are ones that telescope , meaning sums where all but finitely many terms cancel out.
Example 2.
Calculate ∑ n = 3 ∞ 5 4 n 2 − 9 \displaystyle \sum_{n=3}^\infty \frac{5}{4n^2 - 9} n = 3 ∑ ∞ 4 n 2 − 9 5
Solution.
Notice that
1 2 n − 3 − 1 2 n + 3 = 6 4 n 2 − 9 . \frac{1}{2n - 3} - \frac{1}{2n + 3}
= \frac{6}{4n^2 - 9}. 2 n − 3 1 − 2 n + 3 1 = 4 n 2 − 9 6 . So, we get
∑ n = 3 ∞ 5 4 n 2 − 9 = ∑ n = 3 ∞ 5 6 ⋅ 6 4 n 2 − 9 = 5 6 ∑ n = 3 ∞ ( 1 2 n − 3 − 1 2 n + 3 ) . \begin{aligned}
\sum_{n=3}^\infty \frac{5}{4n^2 - 9}
&= \sum_{n=3}^\infty \frac{5}{6} \cdot \frac{6}{4n^2 - 9} \\
&= \frac{5}{6} \sum_{n=3}^\infty \p{\frac{1}{2n - 3} - \frac{1}{2n + 3}}.
\end{aligned} n = 3 ∑ ∞ 4 n 2 − 9 5 = n = 3 ∑ ∞ 6 5 ⋅ 4 n 2 − 9 6 = 6 5 n = 3 ∑ ∞ ( 2 n − 3 1 − 2 n + 3 1 ) . From here, it's always a good idea to look at the partial sums and write out some terms to figure out what cancels out:
∑ n = 3 N ( 1 2 n − 3 − 1 2 n + 3 ) = 1 3 + 1 5 + 1 7 + 1 9 + 1 11 + ⋯ + 1 2 N − 3 − ( 1 9 + 1 11 + ⋯ + 1 2 N − 3 + 1 2 N − 1 + 1 2 N + 1 + 1 2 N + 3 ) = 1 3 + 1 5 + 1 7 − 1 2 N − 1 − 1 2 N + 1 − 1 2 N + 3 . \begin{aligned}
\sum_{n=3}^N \p{\frac{1}{2n - 3} - \frac{1}{2n + 3}}
&= \frac{1}{3} + \frac{1}{5} + \frac{1}{7} + \cancel{\frac{1}{9}} + \bcancel{\frac{1}{11}} + \cdots + \cancel{\frac{1}{2N - 3}} \\
&\qquad - \p{\cancel{\frac{1}{9}} + \bcancel{\frac{1}{11}} + \cdots + \cancel{\frac{1}{2N - 3}} + \frac{1}{2N - 1} + \frac{1}{2N + 1} + \frac{1}{2N + 3}} \\
&= \frac{1}{3} + \frac{1}{5} + \frac{1}{7} - \frac{1}{2N - 1} - \frac{1}{2N + 1} - \frac{1}{2N + 3}.
\end{aligned} n = 3 ∑ N ( 2 n − 3 1 − 2 n + 3 1 ) = 3 1 + 5 1 + 7 1 + 9 1 + 11 1 + ⋯ + 2 N − 3 1 − ( 9 1 + 11 1 + ⋯ + 2 N − 3 1 + 2 N − 1 1 + 2 N + 1 1 + 2 N + 3 1 ) = 3 1 + 5 1 + 7 1 − 2 N − 1 1 − 2 N + 1 1 − 2 N + 3 1 . So, because the infinite sum is (defined to be) the limit of the partial sums,
∑ n = 3 ∞ 5 4 n 2 − 9 = 5 6 lim N → ∞ ∑ n = 3 N ( 1 2 n − 3 − 1 2 n + 3 ) = 5 6 lim N → ∞ ∑ n = 3 N ( 1 3 + 1 5 + 1 7 − 1 2 N − 1 − 1 2 N + 1 − 1 2 N + 3 ) = 5 6 ( 1 3 + 1 5 + 1 7 ) . \begin{aligned}
\sum_{n=3}^\infty \frac{5}{4n^2 - 9}
&= \frac{5}{6} \lim_{N\to\infty} \sum_{n=3}^N \p{\frac{1}{2n - 3} - \frac{1}{2n + 3}} \\
&= \frac{5}{6} \lim_{N\to\infty} \sum_{n=3}^N \p{\frac{1}{3} + \frac{1}{5} + \frac{1}{7} - \frac{1}{2N - 1} - \frac{1}{2N + 1} - \frac{1}{2N + 3}} \\
&= \boxed{\frac{5}{6}\p{\frac{1}{3} + \frac{1}{5} + \frac{1}{7}}}.
\end{aligned} n = 3 ∑ ∞ 4 n 2 − 9 5 = 6 5 N → ∞ lim n = 3 ∑ N ( 2 n − 3 1 − 2 n + 3 1 ) = 6 5 N → ∞ lim n = 3 ∑ N ( 3 1 + 5 1 + 7 1 − 2 N − 1 1 − 2 N + 1 1 − 2 N + 3 1 ) = 6 5 ( 3 1 + 5 1 + 7 1 ) .
Example 3.
Determine whether ∑ n = 1 ∞ ln ( n + 1 n ) \displaystyle \sum_{n=1}^\infty \ln\p{\frac{n + 1}{n}} n = 1 ∑ ∞ ln ( n n + 1 )
Solution.
If you use the properties of logarithms, you get
ln ( n + 1 n ) = ln ( n + 1 ) − ln n . \ln\p{\frac{n+1}{n}} = \ln\p{n+1} - \ln{n}. ln ( n n + 1 ) = ln ( n + 1 ) − ln n . So the partial sums look like
∑ n = 1 N ln ( n + 1 n ) = ∑ n = 1 N ( ln ( n + 1 ) − ln n ) = ln 2 + ln 3 + ⋯ + ln N − ( ln 1 + ln 2 + ln 3 + ⋯ + ln N + ln ( N + 1 ) ) = ln ( N + 1 ) \begin{aligned}
\sum_{n=1}^N \ln\p{\frac{n + 1}{n}}
&= \sum_{n=1}^N \p{\ln\p{n+1} - \ln{n}} \\
&= \cancel{\ln{2}} + \bcancel{\ln{3}} + \cdots + \cancel{\ln{N}} \\
&\quad - \p{\ln{1} + \cancel{\ln{2}} + \bcancel{\ln{3}} + \cdots + \cancel{\ln{N}} + \ln\p{N+1}} \\
&= \ln\p{N + 1}
\end{aligned} n = 1 ∑ N ln ( n n + 1 ) = n = 1 ∑ N ( ln ( n + 1 ) − ln n ) = ln 2 + ln 3 + ⋯ + ln N − ( ln 1 + ln 2 + ln 3 + ⋯ + ln N + ln ( N + 1 ) ) = ln ( N + 1 ) (since ln 1 = 0 \ln{1} = 0 ln 1 = 0
lim N → ∞ ∑ n = 1 N ln ( n + 1 n ) = lim N → ∞ ln ( N + 1 ) = ∞ , \lim_{N\to\infty} \sum_{n=1}^N \ln\p{\frac{n+1}{n}}
= \lim_{N\to\infty} \ln\p{N + 1}
= \infty, N → ∞ lim n = 1 ∑ N ln ( n n + 1 ) = N → ∞ lim ln ( N + 1 ) = ∞ , which means the infinite sum diverges.
Example 4.
Calculate ∑ n = 1 ∞ [ sin 2 ( π 2 n ) − sin 2 ( π 2 n + 2 ) ] \displaystyle \sum_{n=1}^\infty \br{\sin^2\p{\frac{\pi}{2n}} - \sin^2\p{\frac{\pi}{2n + 2}}} n = 1 ∑ ∞ [ sin 2 ( 2 n π ) − sin 2 ( 2 n + 2 π ) ]
Solution.
Like before, we just need to look at the partial sums:
∑ n = 1 N [ sin 2 ( π 2 n ) − sin 2 ( π 2 n + 2 ) ] = sin 2 ( π 2 ) + sin 2 ( π 4 ) + sin 2 ( π 6 ) + ⋯ + sin 2 ( π 2 N ) − [ sin 2 ( π 4 ) + sin 2 ( π 6 ) + ⋯ + sin 2 ( π 2 N ) + sin 2 ( π 2 N + 2 ) ] = 1 − sin 2 ( π 2 N + 2 ) . \begin{aligned}
\sum_{n=1}^N \br{\sin^2\p{\frac{\pi}{2n}} - \sin^2\p{\frac{\pi}{2n + 2}}}
&= \sin^2\p{\frac{\pi}{2}} + \cancel{\sin^2\p{\frac{\pi}{4}}} + \bcancel{\sin^2\p{\frac{\pi}{6}}} + \cdots + \cancel{\sin^2\p{\frac{\pi}{2N}}} \\
&\quad - \br{\cancel{\sin^2\p{\frac{\pi}{4}}} + \bcancel{\sin^2\p{\frac{\pi}{6}}} + \cdots + \cancel{\sin^2\p{\frac{\pi}{2N}}} + \sin^2\p{\frac{\pi}{2N+2}}} \\
&= 1 - \sin^2\p{\frac{\pi}{2N+2}}.
\end{aligned} n = 1 ∑ N [ sin 2 ( 2 n π ) − sin 2 ( 2 n + 2 π ) ] = sin 2 ( 2 π ) + sin 2 ( 4 π ) + sin 2 ( 6 π ) + ⋯ + sin 2 ( 2 N π ) − [ sin 2 ( 4 π ) + sin 2 ( 6 π ) + ⋯ + sin 2 ( 2 N π ) + sin 2 ( 2 N + 2 π ) ] = 1 − sin 2 ( 2 N + 2 π ) . So, when you take limits, you get
lim N → ∞ ∑ n = 1 N [ sin 2 ( π 2 n ) − sin 2 ( π 2 n + 2 ) ] = lim N → ∞ ( 1 − sin 2 ( π 2 N + 2 ) ) = 1 . \lim_{N\to\infty} \sum_{n=1}^N \br{\sin^2\p{\frac{\pi}{2n}} - \sin^2\p{\frac{\pi}{2n + 2}}}
= \lim_{N\to\infty} \p{1 - \sin^2\p{\frac{\pi}{2N+2}}}
= \boxed{1}. N → ∞ lim n = 1 ∑ N [ sin 2 ( 2 n π ) − sin 2 ( 2 n + 2 π ) ] = N → ∞ lim ( 1 − sin 2 ( 2 N + 2 π ) ) = 1 .
Other Problems
Example 5.
For this example, we're going to use the fact that
lim θ → 0 sin θ θ = 1. \lim_{\theta\to0} \frac{\sin\theta}{\theta} = 1. θ → 0 lim θ sin θ = 1. For this limit, what matters is that what's inside the sine and in the denominator are the same, and that they both go to 0 0 0
lim x → 0 sin 2 x x ≠ 1 \lim_{x\to0} \frac{\sin{2x}}{x} \neq 1 x → 0 lim x sin 2 x = 1 since 2 x 2x 2 x x x x 2 2 2
lim x → 0 sin 2 x x = lim x → 0 2 ⋅ sin 2 x 2 x = 2. \lim_{x\to0} \frac{\sin{2x}}{x}
= \lim_{x\to0} 2 \cdot \frac{\sin{2x}}{2x}
= 2. x → 0 lim x sin 2 x = x → 0 lim 2 ⋅ 2 x sin 2 x = 2. With this in mind, without using L'Hôpital's rule, calculate the limit
lim x → 1 sin ( x 2 − 4 x + 3 ) x − 1 . \lim_{x\to1} \frac{\sin\p{x^2 - 4x + 3}}{x - 1}. x → 1 lim x − 1 sin ( x 2 − 4 x + 3 ) .
Solution.
We can factor x 2 − 4 x + 3 = ( x − 1 ) ( x − 3 ) x^2 - 4x + 3 = \p{x - 1}\p{x - 3} x 2 − 4 x + 3 = ( x − 1 ) ( x − 3 )
lim x → 1 sin ( x 2 − 4 x + 3 ) x − 1 = lim x → 1 sin ( ( x − 1 ) ( x − 3 ) ) x − 1 . \lim_{x\to1} \frac{\sin\p{x^2 - 4x + 3}}{x - 1}
= \lim_{x\to1} \frac{\sin\p{\p{x - 1}\p{x - 3}}}{x - 1}. x → 1 lim x − 1 sin ( x 2 − 4 x + 3 ) = x → 1 lim x − 1 sin ( ( x − 1 ) ( x − 3 ) ) . Finally, we just need to make sure what's inside the sine and what's in the denominator are the same, so we can multiply and divide by x − 3 x - 3 x − 3
lim x → 1 sin ( ( x − 1 ) ( x − 3 ) ) ( x − 1 ) ( x − 3 ) ⋅ ( x − 3 ) = − 2 . \lim_{x\to1} \frac{\sin\p{\p{x - 1}\p{x - 3}}}{\p{x - 1}\p{x - 3}} \cdot \p{x - 3}
= \boxed{-2}. x → 1 lim ( x − 1 ) ( x − 3 ) sin ( ( x − 1 ) ( x − 3 ) ) ⋅ ( x − 3 ) = − 2 .
Example 6.
Calculate ∫ tan 2 x d x \displaystyle \int \tan^2{x} \,\diff{x} ∫ tan 2 x d x
Solution.
If you use the identity tan 2 x + 1 = sec 2 x \tan^2{x} + 1 = \sec^2{x} tan 2 x + 1 = sec 2 x
∫ tan 2 x d x = ∫ sec 2 x − 1 d x = tan x − x + C . \int \tan^2{x} \,\diff{x}
= \int \sec^2{x} - 1 \,\diff{x}
= \boxed{\tan{x} - x + C}. ∫ tan 2 x d x = ∫ sec 2 x − 1 d x = tan x − x + C .