Esda tutingki, biz ketma-ket atamalarni "o'chirib tashlaymiz" deganimizda, biz ulardan qutula olmaymiz. Biz qilayotgan barcha narsa qatorning birinchi bir necha shartlarini ketma-ketlik oldiga yig'indi sifatida yozishdir.

Shunday qilib, ushbu seriya uchun dastlabki uchta atamani olib tashlash,

\[\begin{align*}\sum\limits_{n = 1}^\infty {\frac{{{2^{ - n}}}}{{{n^2} + 1}}} & = \frac{{{2^{ - 1}}}}{{{1^2} + 1}} + \frac{{{2^{ - 2}}}}{{{2^2} + 1}} + \frac{{{2^{ - 3}}}}{{{3^2} + 1}} + \sum\limits_{n = 4}^\infty {\frac{{{2^{ - n}}}}{{{n^2} + 1}}} \\ & = \frac{1}{4} + \frac{1}{{20}} + \frac{1}{{80}} + \sum\limits_{n = 4}^\infty {\frac{{{2^{ - n}}}}{{{n^2} + 1}}} \\ & = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{5}{{16}} + \sum\limits_{n = 4}^\infty {\frac{{{2^{ - n}}}}{{{n^2} + 1}}} }}\end{align*}\]

Bu birinchi qadam unchalik zarur emas, lekin biz \(n = 1\), \(n = 2\) va \(n = 3\) (i.e. \(n\) ning dastlabki uchta qiymati) umumiy ketma-ketlik terminiga kiritiladi. Bundan tashqari, \(n\) ning boshlang‘ich qiymatini o‘zgartirishni unutmang, chunki biz \(n\) yoki atamalarning dastlabki uchta qiymatini “o‘chirib tashlagan”miz.