Aquileo | Sequence & Series PYQs

In examinations, questions from sequence and series are frequently asked in the form of finding the nth term, convergence or divergence, as well as problems based on special series formulas and tests of convergence.

Short Questions on Sequence & Series

Question 1: Test the convergence of the series: \sum_{n=1}^{\infty} \frac{1}{n^2} + \frac{1}{n+1^{2}}.

Question 2: Test the convergence of the series f(x) = \dfrac{n!}{3^n}.

Question 3: Test of the coverage of the following series

2/3 + 2.4/3.5 + 2.4.6/3.5.7 + 2.4.6.8/3.5.7.9 + ...

Question 5: Find the Fourier series of f(x) = x in [−π,π].

Question 6: For the function f(x) = cos⁡(4x), find the fundamental period.

Question 7: Find the sum of the geometric series S = 3 + 6 + 12 +… up to 8 terms.

Question 8: Determine whether the function f(t) = sin⁡(3t) + cos⁡(5t) is periodic. If yes, find its fundamental period.

Question 9: Find the fundamental period of f(t) = cos⁡(t/2) + sin⁡(t/3).

Question 10: Express f(x)=x⁵+4x²+ 7 as a sum of even and odd parts.

Long Type Questions on Sequence & Series

Question 1: Test the convergence of the series \sum_{n=1}^{\infty} \frac{n}{n^2+1}.n

Question 2: Show that the function f(x) = sin⁡(3x) + cos⁡(5x) is periodic and find its fundamental period.

Check if you were right - full answer with solution below.

Short Question onSequence & Series: Answers

Solution 1:

Here u_n = [\frac{1}{n^2} + \frac{1}{n+1^{2}}]
Clearly \frac{1}{n^2} + \frac{1}{n+1^{2}} < \frac{1}{n^2} + \frac{1}{n^{2}} < \frac{1}{n^2}
Now \sum_{n=1}^{\infty} \frac{1}{n^2} is a covergent series ( As p = 2 > 1).
Thus by comparsion test \sum_{n=1}^{\infty} \frac{1}{n^2} + \frac{1}{n+1^{2}} is also covergent.

Solution 2:

Let a_n = \dfrac{n!}{3^n}. Use the Ratio Test: compute
L= \lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right| = \lim_{n\to\infty}\frac{(n+1)!/3^{\,n+1}}{n!/3^n} = \lim_{n\to\infty}\frac{n+1}{3}.
Evaluate the limit: \dfrac{n+1}{3}→∞ as n→∞. So L = ∞ > 1 .
Ratio Test conclusion: if L > 1 (or infinite), the series diverges.

Solution 3:

Here u_n = \frac{2}{3} + \frac{2 \cdot 4}{3 \cdot 5} + \frac{2 \cdot 4 \cdot 6}{3 \cdot 5 \cdot 7} + \frac{2 \cdot 4 \cdot 6 \cdot 8}{3 \cdot 5 \cdot 7 \cdot 9} + \cdots
= u_{n +1}= = \sum_{n=1}^{\infty} \frac{2 \cdot 4 \cdot 6 \cdot \cdots \cdot (2n)}{3 \cdot 5 \cdot 7 \cdot \cdots \cdot (2n+1)}
Then
= \sum_{n=1}^{\infty} \frac{(2n)!!}{(2n+1)!!}

Solution 4:

Split into even/odd parts:
x⁴ → even → \int_{-2}^{2} x^4 dx = 2 \int_0^2 x^4 dx = 2 \cdot \frac{2^5}{5} = \frac{64}{5}
x³→ odd → \int_{-2}^{2} 3x^3 dx = 0
2 → even → \int_{-2}^{2} 2 dx = 2. 2 = 4
Total = 64/5 + 0 + 4 = 64/4 + 20/5 = 84/5

Solution 5:

Since f(x) = x is an odd function, Fourier series contains only sine terms.
a₀= 0
a_n= 0
b_n = \frac{1}{\pi}\int_{-\pi}^{\pi} x\sin(nx) \ dx= \frac{2}{\pi}\int_{0}^{\pi} x\sin(nx) \ dx
Evaluate the integral:
\int x\sin(nx)\ dx= -\frac{x\cos(nx)}{n} + \frac{\sin(nx)}{n^2}
So, \int_{0}^{\pi} x\sin(nx)\ dx= \left[-\frac{x\cos(nx)}{n} + \frac{\sin(nx)}{n^2}\right]_0^{\pi}= -\frac{\pi\cos(n\pi)}{n}= \frac{\pi(-1)^{,n+1}}{n}
Hence b_n = \frac{2}{\pi}\cdot \frac{\pi(-1)^{,n+1}}{n}= 2,\frac{(-1)^{,n+1}}{n}.
Fourier series:
x = 2 \sum_{n=1}^\infty (-1)^{n+1} \frac{\sin(nx)}{n}x

Solution 6:

f(x) = cos(4x)
The general period of cos⁡(kx) is: T= 2π/k
For f(x) = cos⁡(4x), k = 4, so:T = 2π/4=π/2
This means the function repeats its values every π/2 units.

Solution 7:

The first term is
a = 3
The common ratio is
r= 6/3 = 2
The formula for the sum of the first n terms of a geometric series is
S_n= a⋅rⁿ− 1/r−1, r ≠ 1
Substitute a=3,r = 2, and n = 8:
S₈=3 \cdot \frac{2^8 - 1}{2 - 1}
S₈= 3⋅255
S₈=765

Solution 8:

f(t) = sin⁡(3t) + cos⁡(5t)
The period of sin⁡(3t) is
T₁= 2π/3
The period of cos⁡(5t) is
T₂= 2π/5
The fundamental period is the LCM of T₁ and T₂:
T = LCM(2π/3,2π/5)
T = LCM(2π/3 ,2π/5)
T = 2π/gcd⁡(3,5) = 2π/1 = 2π

Solution 9:

Period of cos⁡(t/2):
T₁= 2π.(1/2) = 4π
Period of sin⁡(t/3):
T₂= 2π.(1/3) = 6
Fundamental period = LCM(4π,6π)
T = 12π

Solution 10:

f_{\text{even}}(x) = \frac{f(x) + f(-x)}{2}, \quad f_{\text{odd}}(x) = \frac{f(x) - f(-x)}{2}
(−x)⁵+ 4(−x)²+ 7 = −x⁵+ 4x²+ 7
f_even(x) = (x⁵+ 4x²+ 7) + (−x⁵+ 4x²+ 7)/2
= \frac{(x^5 + 4x^2 + 7) + (-x^5 + 4x^2 + 7)}{2}= \frac{2(4x^2 + 7)}{2}= 4x^2 + 7
Answer: Even part = (4x²+7) + x⁵, Odd part = x⁵

Long Question on Sequence & Series: Answers

Solution 1:

Let
a_n = \frac{n}{n^2+1}
For large n,
a_n ≈ n/n²= 1/n
Compare with the harmonic series:
∑1/n which diverges.
Using Limit Comparison Test with b_n= 1/n:
\lim_{n \to \infty} \frac{a_n}{b_n}= \lim_{n \to \infty} \frac{\tfrac{n}{n^2+1}}{\tfrac{1}{n}}= \lim_{n \to \infty} \frac{n^2}{n^2+1}= 1.
Since the limit is finite and non-zero, and ∑1/n diverges,
\sum_{n=1}^{\infty} \frac{n}{n^2+1}

Solution 2:

Period of sin⁡(3x) is
T₁= 2π/3
Period of cos⁡(5x) is
T₂= 2π/5
For the sum to be periodic, the fundamental period is the LCM of individual periods:
T = \text{LCM}\left(\frac{2\pi}{3}, \frac{2\pi}{5}\right).
Since denominators are 3 and 5,
T = 2π/gcd⁡(3, 5)⋅15 = 2π×15/15
Thus,
f(x) is periodic with fundamental period 2π.b_n = \frac{1}{\pi}\int_{-\pi}^{\pi} x\sin(nx)\ dx= \frac{2}{\pi}\int_{0}^{\pi} x\sin(nx) \ dx

Sequence & Series PYQs

Short Questions on Sequence & Series

Long Type Questions on Sequence & Series

Check if you were right - full answer with solution below.

Short Question onSequence & Series: Answers

Long Question on Sequence & Series: Answers

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