sum of 100 terms of logarithmic expression

Question

Calculate value of

$\displaystyle \sum^{100}_{k=1}\ln\bigg(\frac{(2k+1)^4+\frac{1}{4}}{16k^4+\frac{1}{4}}\bigg)$

My try :: $\displaystyle x^4+4y^4$

$=(x^2+2xy+2y^2)(x^2-2xy+2y^2)$

So sum $\displaystyle \sum^{100}_{k=1}\ln\bigg(\frac{1+4(2k+1)^4}{1+4(2k)^4}\bigg)$

$\displaystyle =\sum^{100}_{k=1}\ln\bigg[\frac{(1+2(2k+1)+2(2k+1)^2)(1-2(2k+1)+2(2k+1)^2)}{(1+2(2k)+2(2k)^2)(1-2(2k)+2(2k)^2)}\bigg]$

How can I decompose that complex expression into partial fractions?

Answer 1

$$g(k)=\dfrac{(2k+1)^4+\dfrac14}{(2k)^4+\dfrac14}=\dfrac{(4k+2)^4+4\cdot1^4}{(4k)^4+4\cdot1^4}$$

Using your formula only,

$\displaystyle x^4+4y^4=(x^2+2y^2)^2-(2xy)^2=(x^2+2xy+2y^2)(x^2-2xy+2y^2)$

$x=4k+2, y=1$

$\displaystyle\implies x^2+2xy+2y^2=(4k+2)^2+2(4k+2)+2=16k^2+24k+10\ \ \ \ (1)$

$x^2-2xy+2y^2=(4k+2)^2-2(4k+2)+2=16k^2+8k+2\ \ \ \ (2)$

$x=4k, y=1$

$\displaystyle\implies x^2+2xy+2y^2=(4k)^2+2(4k)+2=?\ \ \ \ (3)$ which cancels $(2)$

$\displaystyle\implies x^2-2xy+2y^2=(4k)^2-2(4k)+2=16k^2-8k+2=(4k-2)^2+2(4k-2)+2\ \ \ \ (4)$

$$\implies\prod_{k=1}^n\dfrac{(2k+1)^4+\dfrac14}{(2k)^4+\dfrac14}=\prod_{k=1}^n\dfrac{(4k+2)^2+2(4k+2)+2}{(4(k-1)+2)^2+2(4(k-1)+2)+2}=\prod_{k=1}^n\dfrac{f(k)}{f(k-1)}$$ where $f(m)=(4m+2)^2+2(4m+2)+2$

$$\implies\prod_{k=1}^n\dfrac{(2k+1)^4+\dfrac14}{(2k)^4+\dfrac14}=\dfrac{f(n)}{f(0)}$$

Answer 2

$\ln (16241)$. Work out the first few cases, use $\ln a + \ln b = \ln ab$ and then prove by induction.

Answer 3

Let's do bit by bit. $$\frac{(2k+1)^4+\frac{1}{4}}{16k^4+\frac{1}{4}}$$ $$=\frac{(k+\frac{1}{2})^4+\frac{1}{64}}{k^4+\frac{1}{64}}$$ Now, $$\ln\Big( \frac{(k+\frac{1}{2})^4+\frac{1}{64}}{k^4+\frac{1}{64}} \Big)$$ $$=\ln\Big( (k+\frac{1}{2})^4+\frac{1}{64} \Big)-\ln\Big( k^4+\frac{1}{64} \Big)$$

Now bring the summation. $$\sum^{100}_{k=1} \ln\Big( (k+\frac{1}{2})^4+\frac{1}{64} \Big)-\ln\Big( k^4+\frac{1}{64} \Big)$$

$$=\Big[\ln\Big( (1+\frac{1}{2})^4+\frac{1}{64} \Big) - \ln\Big( 1^4+\frac{1}{64} \Big)\Big] + \Big[ \ln\Big( (2+\frac{1}{2})^4+\frac{1}{64} \Big) - \ln\Big( 2^4+\frac{1}{64} \Big) \Big] + \Big[ \ln\Big( (3+\frac{1}{2})^4+\frac{1}{64} \Big) - \ln\Big( 3^4+\frac{1}{64} \Big) \Big] + \dots + \Big[ \ln\Big( (100+\frac{1}{2})^4+\frac{1}{64} \Big) - \ln\Big( 100^4+\frac{1}{64} \Big) \Big]$$

Hope you can continue after this as it's now simplified